cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A122485 Values of A083097(k) such that A083097(k) = A083097(k+1) - 1.

Original entry on oeis.org

5, 14, 41, 59, 122, 140, 167, 176, 365, 383, 410, 419, 491, 500, 527, 545, 1094, 1112, 1139, 1148, 1220, 1229, 1256, 1274, 1463, 1472, 1499, 1517, 1580, 1598, 1625, 1634, 3281, 3299, 3326, 3335, 3407, 3416, 3443, 3461, 3650, 3659, 3686, 3704, 3767, 3785, 3812
Offset: 1

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Author

Alexander Adamchuk, Sep 15 2006

Keywords

Comments

A083097(n) = A083095(n) = A083096(n)/6 = A083094(n)/4, where A083096 are the Numbers k such that 3 divides Sum_{j=1..k} C(2*j,j) = A066796(k).
All terms are of the form 9*m + 5 and belong to A017221 with m = {0, 1, 4, 6, 13, 15, 18, 19, 40, 42, ...}.
Corresponding numbers m such that a(m) = A083097(m) are A129771 (evil odd numbers).

Examples

			A083097 begins {0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, ...}.
So a(1) = 5 because 5 = A083097(3) = A083097(3+1) - 1.
a(2) = 14 because 14 = A083097(5) = A083097(5+1) - 1.
		

Crossrefs

Formula

a(n) = A083097(A129771(n)).

Extensions

More terms from R. J. Mathar, Jan 17 2008
More terms from Jinyuan Wang, Jan 22 2022
Edited by Michel Marcus, Jan 22 2022

A083096 Numbers k such that 3 divides Sum_{j=1..k} binomial(2*j,j).

Original entry on oeis.org

0, 12, 30, 36, 84, 90, 108, 120, 246, 252, 270, 282, 324, 336, 354, 360, 732, 738, 756, 768, 810, 822, 840, 846, 972, 984, 1002, 1008, 1056, 1062, 1080, 1092, 2190, 2196, 2214, 2226, 2268, 2280, 2298, 2304, 2430, 2442, 2460, 2466, 2514, 2520, 2538, 2550
Offset: 1

Views

Author

Benoit Cloitre, Apr 22 2003

Keywords

Comments

Apparently a(n)/3 (mod 3) = A010060(n-1), the Thue-Morse sequence.

Crossrefs

Programs

Formula

It appears that sequence gives k such that the coefficient of x^k equals 1 in Product_{j>=1} 1-x^(3^j).

A083095 a(n) = A083094(n)/4.

Original entry on oeis.org

0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, 122, 123, 126, 128, 135, 137, 140, 141, 162, 164, 167, 168, 176, 177, 180, 182, 365, 366, 369, 371, 378, 380, 383, 384, 405, 407, 410, 411, 419, 420, 423, 425, 486, 488, 491, 492, 500
Offset: 1

Views

Author

Benoit Cloitre, Apr 22 2003

Keywords

Comments

Is this the same as A083097? - Andrew S. Plewe, May 30 2007

Crossrefs

Programs

  • Mathematica
    Select[Range[0,2000], OddQ[Sum[Mod[Binomial[#,j],3],{j,0,#}]]&]/4 (* Paul F. Marrero Romero, Dec 28 2024 *)
  • Python
    def A083095(n): return int(bin(((m:=n-1).bit_count()&1)+(m<<1))[2:],3)>>1 # Chai Wah Wu, Jun 26 2025

Formula

Apparently (2*a(n)) mod 3 = A010060(n-1), the Thue-Morse sequence.
Numbers k such that C(4*k, 2*k) == 1 (mod 3). - Benoit Cloitre, Jul 30 2003
Numbers k such that the base-3 digits of 2k contains no 2's, i.e. a(n) = A074939(n-1)/2. - Chai Wah Wu, Jun 26 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 29 2003

A167912 a(n) = (1/(3^n)^2) * Sum_{k=0..(3^n-1)} binomial(2k,k).

Original entry on oeis.org

1, 217, 913083596083, 18744974860247264575032720770000376335095039
Offset: 1

Views

Author

Alexander Adamchuk, Nov 15 2009

Keywords

Comments

Note that a(n) mod 27 = a(n) mod 9 = a(n) mod 3 = 1.
The Maple program yields the first seven terms; easily adjustable for obtaining more terms. However, a(4) has 44 digits, a(5) has 140 digits, a(6) has 432 digits and a(7) has 1308 digits. - Emeric Deutsch, Nov 22 2009

Crossrefs

Programs

  • Maple
    a := proc (n) options operator, arrow: (sum(binomial(2*k, k), k = 0 .. 3^n-1))/3^(2*n) end proc: seq(a(n), n = 1 .. 7); # Emeric Deutsch, Nov 22 2009
  • Mathematica
    Table[(1/3^n)^2 * Sum[Binomial[2 k, k], {k, 0, 3^n - 1}], {n, 1, 5}] (* G. C. Greubel, Jul 01 2016 *)

Extensions

a(4) from Emeric Deutsch, Nov 22 2009

A181990 a(n) = Sum_{0 <= k <= m < p} (binomial(m, k)^(p-1))/p, where p is the n-th prime.

Original entry on oeis.org

3, 399, 12708885, 124515078454872901983423, 39212583445587381894247266262023061, 43487633454143579523135045521112077473364484383507327790688372131, 157851796824901989964381293031623545741924564754192453966085327785455257503133278729
Offset: 2

Views

Author

Alexander Adamchuk, Apr 04 2012

Keywords

Comments

a(n) is a sum of all elements in the first p rows of Pascal's triangle each raised to the (p-1) power and divided by p, where p is the n-th prime.
For p = 3 and 7 (and their powers like 3, 9, 27, ... and 7, 49, ...) the sums of all elements in n = p^k top rows of Pascal's triangle each raised to the (n-1) = (p^k-1) power are divisible by n^2 = p^(2k) for all k > 0.

Crossrefs

Programs

  • Mathematica
    Table[(Sum[Binomial[m, k]^(Prime[n] - 1), {m, 0, Prime[n] - 1}, {k, 0, m}])/Prime[n], {n, 2, 10}]
  • PARI
    a(n) = my(p=prime(n)); sum(m=0, p-1, sum(k=0, m, binomial(m,k)^(p-1))/p); \\ Michel Marcus, Dec 03 2018
Showing 1-5 of 5 results.