The On-Line Encyclopedia of Integer Sequences (OEIS)

Revisions by M. F. Hasler

(See also M. F. Hasler's wiki page
and changes approved by M. F. Hasler)

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A081357

Sublime numbers, numbers for which the number of divisors and the sum of the divisors are both perfect.
(history; published version)

#37 by M. F. Hasler at Wed Mar 05 17:51:57 EST 2025

STATUS

editing

proposed

#36 by M. F. Hasler at Wed Mar 05 17:49:53 EST 2025

COMMENTS

It is known that all sublime numbers must be of that form, with all odd prime factors being Mersenne primes M(p). - M. F. Hasler, Mar 05 2025

STATUS

proposed

editing

Discussion

Wed Mar 05

17:51

M. F. Hasler: The second comment could also go into the example section. In that case I would reword mine a little.

A061808

a(n) is the smallest number with all digits odd that is divisible by 2n-1.
(history; published version)

#56 by M. F. Hasler at Wed Mar 05 11:03:17 EST 2025

STATUS

editing

proposed

#55 by M. F. Hasler at Wed Mar 05 11:03:12 EST 2025

DATA

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 315, 115, 75, 135, 319, 31, 33, 35, 37, 39, 533, 559, 135, 517, 539, 51, 53, 55, 57, 59, 793, 315, 195, 335, 759, 71, 73, 75, 77, 79, 1377, 913, 595, 957, 979, 91, 93, 95, 97, 99, 1111, 515, 315, 535, 1199, 111, 113, 115, 117, 119, 1331, 1353, 375, 1397, 1935

COMMENTS

All terms must be odd. - M. F. Hasler, Mar 05 2025

FORMULA

a(n) = (2n-1)*A296009(n). - _From _M. F. Hasler_, Mar 05 2025: (Start)

a(n) = (2n-1)*A296009(n).

a(n) == 1 (mod 2) for all n. (End)

PROG

(PARI) apply( {A061808(n)=forstep(k=n*2-1, oo, n*4-2, vecmin(digits(k)%2)&& return(k))}, [1..99])

(Python) A061808 = lambda n: next(m for m in range(2*n-1, 9<<99, 4*n-2) if all(int(d)&1 for d in str(m))) # M. F. Hasler, Mar 05 2025

STATUS

proposed

editing

#54 by M. F. Hasler at Wed Mar 05 10:50:42 EST 2025

STATUS

editing

proposed

#53 by M. F. Hasler at Wed Mar 05 10:50:05 EST 2025

COMMENTS

From Yang Haoran, Dec 02 2017, edited by _M. F. Hasler_, Mar 05 2025: (Start)

Record value for a(n) = (2n-1) * A296009(n).:

(the list above is complete up to here)

...

990001 * 12121113 = 11999913991113 (the first A296009(n) > 2n-1).

Discussion

Wed Mar 05

10:50

M. F. Hasler: tentatively edited ; makes much more sense now, IMO.

#52 by M. F. Hasler at Wed Mar 05 10:44:16 EST 2025

FORMULA

a(n) = (2n-1)*A296009(n). - M. F. Hasler, Mar 05 2025

Discussion

Wed Mar 05

10:48

M. F. Hasler: I don't understand the COMMENTS: 
* "Record value for a(n)=...(PERIOD)" ?!? do they mean ":" ?
* "The above list is complete." ??? 
(1) It's obviously not; it's truncated. 
(2) it wouldn't make sense (and not be the list of records) if values were omitted.
So, maybe at least "complete up to here" ?

#51 by M. F. Hasler at Wed Mar 05 10:43:21 EST 2025

LINKS

Mathematical Excalibur, <a href="https://www.math.ust.hk/excalibur/v13_n1.pdf"> Problem 300</a>, Vol. 1 No. 3 p. 3, (2008), p. 3.

STATUS

approved

editing

A381699

allocated for Ali Sada
(history; published version)

#21 by M. F. Hasler at Wed Mar 05 10:39:16 EST 2025

STATUS

editing

proposed

#20 by M. F. Hasler at Wed Mar 05 10:39:09 EST 2025

COMMENTS

Records values of k(n) = a(n)/2n are k(1) = 2, k(5) = 3, k(7) = 4, k(11) = 5, k(21) = 8, k(45) = 11, k(58) = 12, k(101) = 55, k(182) = 108, k(1001) = 555, k(2001) = 778, k(3996) = 1001, k(7992) = 3253, k(9091) = 21545, k(9901) = 161155, ...