cp's OEIS Frontend

These are Hogben's central polygonal numbers

2

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3 n

Also the sum of three consecutive triangular numbers (A000217); i.e., a(4) = 19 = T4 + T3 + T2 = 10 + 6 + 3. - Robert G. Wilson v, Apr 27 2001

For k>2, Sum_{n=1..k} a(n) gives the sum pertaining to the magic square of order k. E.g., Sum_{n=1..5} a(n) = 1 + 4 + 10 + 19 + 31 = 65. In general, Sum_{n=1..k} a(n) = k*(k^2 + 1)/2. - Amarnath Murthy, Dec 22 2001

Binomial transform of (1,3,3,0,0,0,...). - Paul Barry, Jul 01 2003

a(n) is the difference of two tetrahedral (or pyramidal) numbers: C(n+3,3) = (n+1)(n+2)(n+3)/6. a(n) = A000292(n) - A000292(n-3) = (n+1)(n+2)(n+3)/6 - (n-2)(n-1)(n)/6. - Alexander Adamchuk, May 20 2006

Partial sums are A006003(n) = n(n^2+1)/2. Finite differences are a(n+1) - a(n) = A008585(n) = 3n. - Alexander Adamchuk, Jun 03 2006

If X is an n-set and Y a fixed 3-subset of X then a(n-2) is equal to the number of 3-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007

Equals (1, 2, 3, ...) convolved with (1, 2, 3, 3, 3, ...). a(4) = 19 = (1, 2, 3, 4) dot (3, 3, 2, 1) = (3 + 6 + 6 + 4). - Gary W. Adamson, May 01 2009

Equals the triangular numbers convolved with [1, 1, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009

a(n) is the number of triples (w,x,y) having all terms in {0,...,n} and min(w+x,x+y,y+w) = max(w,x,y). - Clark Kimberling, Jun 14 2012

a(n) = number of atoms at graph distance <= n from an atom in the graphite or graphene network (cf. A008486). - N. J. A. Sloane, Jan 06 2013

In 1826, Shiraishi gave a solution to the Diophantine equation a^3 + b^3 + c^3 = d^3 with b = a(n) for n > 1; see A226903. - Jonathan Sondow, Jun 22 2013

For n > 1, a(n) is the remainder of n^2 * (n-1)^2 mod (n^2 + (n-1)^2). - J. M. Bergot, Jun 27 2013

The equation A000578(x) - A000578(x-1) = A000217(y) - A000217(y-2) is satisfied by y=a(x). - Bruno Berselli, Feb 19 2014

A242357(a(n)) = n. - Reinhard Zumkeller, May 11 2014

A255437(a(n)) = 1. - Reinhard Zumkeller, Mar 23 2015

The first differences give A008486. a(n) seems to give the total number of triangles in the n-th generation of the six patterns of triangle expansion shown in the link. - Kival Ngaokrajang, Sep 12 2015

Number of binary shuffle squares of length 2n which contains exactly two 1's. - Bartlomiej Pawlik, Sep 07 2023

The digital root has period 3 (1, 4, 1) (A146325), the same digital root as the centered 12-gonal numbers, or centered dodecagonal numbers A003154(n). - Peter M. Chema, Dec 20 2023

Numbers w such that w^3 = x^3+y^3+z^3, x>y>z>=0, is soluble.

A226903(n) + 1 is an infinite subsequence parametrized by Shiraishi in 1826. - Jonathan Sondow, Jun 22 2013

Because of Fermat's Last Theorem, sequence lists numbers w such that w^3 = x^3+y^3+z^3, x>y>z>0, is soluble. In other words, z cannot be 0 because x and y are positive integers by definition of this sequence. - Altug Alkan, May 08 2016

This sequence is the same as numbers w such that w^3 = x^3+y^3+z^3, x>=y>=z>0, is soluble as Legendre showed that a^3+b^3=2*c^3 only has the trivial solutions a = b or a = -b (see Dickson's History of the Theory of Numbers, vol. II, p. 573). - Chai Wah Wu, May 13 2017

Solutions x to the equations x = a^3 + b^3 = c^3 - d^3 in positive integers.

The intersection of A003325 and A181123. See those sequences for additional comments, references, links and cross-refs.

Suggested by Shiraishi's solutions to Gokai Ampon's equation u^3 + v^3 + w^3 = n^3 (transpose a term from the left side to the right side). See A023042 and A226903.

An infinite subsequence is (A226904(n)+1)^3 - A226904(n)^3.

The numbers c in A225909.

The sequence is infinite, because A226903 is a parametrized infinite subsequence.

Solutions x to the equations x = a^3 + b^3 = (c+1)^3 - c^3 in positive integers. The values of c are A226902.

The intersection of A003325 and A003215.

Subsequence of A225908 = numbers that are both a sum and a difference of two positive cubes.

Shiraishi's solution to Gokai Ampon's equation u^3 + v^3 + w^3 = n^3 (see A023042 and A226903) shows that the sequence is infinite.

Complement of A226903 in A226902.

The data are derived from the following formula:

(a^3 - 6*t^3)^3 + (a^3 + 6*t^3)^3 + (-6*a*t^2)^3 = 2*a^9;

(4*a^3 - 3*t^3)^3 + (4*a^3 + 3*t^3)^3 + (-6*a*t^2)^3 = 128*a^9 = 2*4^3*a^9;

(9*a^3 - 2*t^3)^3 + (9*a^3 + 2*t^3)^3 + (-6*a*t^2)^3 = 1458*a^9 = 2*9^3*a^9;

(36*a^3 - t^3)^3 + (36*a^3 + t^3)^3 + (-6*a*t^2)^3 = 93312*a^9 = 2*36^3*a^9;

((3*a^3)*t - 9*t^4)^3 + (9*t^4)^3 + (a^4 - 9*a*t^3)^3 = a^12;

((9*a^3)*t - t^4)^3 + (t^4)^3 + (9*a^4 - 3*a*t^3)^3 = 729*a^12 = 9^3*a^12.

The data are derived from the following formula:

(a^2 - a*t - t^2)^3 + (a^2 + a*t - t^2)^3 + 2*(t^2)^3 = 2*a^6

(a^3 - 3*t^3)^3 + (a^3 + 3*t^3) + 2*(-3*a*t^2)^3 = 2*a^9;

(9*a^3 - t^3)^3 + (9*a^3 + t^3)^3 + 2*(-3*a*t^2)^3 = 1458*a^9;

(6*a^3*t - 72*t^4)^3 + (72*t^4)^3 + 2*(a^4 - 36*a*t^3)^3 = 2*a^12;

(6*a^3*t - 9*t^4)^3 + (9*t^4)^3 + 2*(2*a^4 - 9*a*t^3)^3 = 16*a^12 = 2*2^3*a^12;

(18*a^3*t - 8*t^4)^3 + (8*t^4)^3 + 2*(9*a^4 - 12*a*t^3)^3 = 1458*a^12 = 2*9^3*a^12;

(18*a^3*t - t^4)^3 + (t^4)^3 + 2*(18*a^4 - 3*a*t^3)^3 = 11664*a^12 = 2*18^3*a^12.

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.

The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.