The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A103532 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A103532

Number of divisors of 240^n.
(history; published version)

#53 by Charles R Greathouse IV at Thu Sep 08 08:45:17 EDT 2022

PROG

(MAGMAMagma) [(4*n+1)*(n+1)^2: n in [0..45]]; // Vincenzo Librandi, Feb 10 2016

Discussion

Thu Sep 08

08:45

OEIS Server: https://oeis.org/edit/global/2944

#52 by Joerg Arndt at Sat Oct 27 03:21:55 EDT 2018

FORMULA

Starting from the Binomial Develop for n^3= Sum_{X=1}^{n} (3*X^2-3*X+1), stretching Y by changing: X=(2x-1): a(n)=Sum_{x=1}^{n} {3*(2x-1)^2-3*(2x-1)+1} [Stefano Maruelli, Oct 11 2018]

STATUS

editing

approved

#51 by Stefano Maruelli at Thu Oct 11 05:23:36 EDT 2018

FORMULA

Starting from the Binomial Develop for n^3= Sum_{X=1}^{n} (3*nX^2-3*nX+1), stretching Y by changing: xX=(2X2x-1): a(n)=Sum_{x=1}^{n} {3*(2n2x-1)^2-3*(2n2x-1)+1. } [Stefano Maruelli, Oct 11 2018]

Discussion

Thu Oct 11

11:55

Joerg Arndt: This is very confusing and doesn't even seem to belong here.

Thu Oct 25

19:16

OEIS Server: This sequence has not been edited or commented on for a week
yet is not proposed for review.  If it is ready for review, please
visit https://oeis.org/draft/A103532 and click the button that reads
"These changes are ready for review by an OEIS Editor."

Thanks.
  - The OEIS Server

#50 by Michel Marcus at Thu Oct 11 02:56:50 EDT 2018

STATUS

proposed

editing

Discussion

Thu Oct 11

05:20

Stefano Maruelli: Sorry as told in this way will be totally missed the scope of the notation: as you can see in this way this can be sticked on the cartesian plane and, most important it shows that is just one of infinite mny other modification you can apply using this Stretch technique... I know is not "traditional", but it will surprise many readers of how powerfull is this new instrument. Pls visit: http://maruelli.com/two-hand-clock/Maruelli-The two Hands Clock Vol.1-2018-04.pdf

#49 by Michel Marcus at Thu Oct 11 01:38:16 EDT 2018

STATUS

editing

proposed

#48 by Michel Marcus at Thu Oct 11 01:37:22 EDT 2018

FORMULA

Starting from the Binomial Develop for n^3= Sum_{1}^{n} (3*n^2-3*n+1) Stretching , stretching Y by changing: x=(2X-1): a(n)=Sum_{1}^{n} 3*(2n-1)^2-3*(2n-1)+1 . [Stefano Maruelli, Oct 11 2018]

STATUS

proposed

editing

Discussion

Thu Oct 11

01:38

Michel Marcus: please can you change the sums to look like a(n) = Sum_{i=0..n} (n+1)*(8*i+1). (2 formulas before yours)

#47 by Stefano Maruelli at Thu Oct 11 01:26:25 EDT 2018

STATUS

editing

proposed

#46 by Stefano Maruelli at Thu Oct 11 00:46:49 EDT 2018

FORMULA

Starting from the Binomial Develop for n^3= Sum_{1}^{n} (3*n^2-3*n+1) Stretching Y by changing: x=(2X-1): a(n)=Sum_{1}^{n} 3*(2n-1)^2-3*(2n-1)+1 [Stefano Maruelli, Oct 11 2018]

#45 by Stefano Maruelli at Thu Oct 11 00:39:31 EDT 2018

FORMULA

Starting from the Binomial Develop for n^3= Sum_{1}^{n} (3*n^2-3*n+1) Stretching Y by changing: x=(2X-1): Sum_{1}^{n} 3*(2n-1)^2-3*(2n-1)+1 [Stefano Maruelli, Oct 11 2018]

STATUS

approved

editing

Discussion

Thu Oct 11

00:41

Stefano Maruelli: There is an important connection here to what is known as Umbral calculus, and what I've called complicate modulus algebra: starting from the Binomial Develop for n^3= Sum_{1}^{n} (3*n^2-3*n+1) Stretching Y by changing: x=(2X-1) we get this sequence as: a(n)= Sum_{1}^{n} 3*(2n-1)^2-3*(2n-1)+1 .

#44 by Vaclav Kotesovec at Tue Oct 04 13:59:04 EDT 2016

STATUS

editing

approved