A213772 Principal diagonal of the convolution array A213771.

1, 11, 42, 106, 215, 381, 616, 932, 1341, 1855, 2486, 3246, 4147, 5201, 6420, 7816, 9401, 11187, 13186, 15410, 17871, 20581, 23552, 26796, 30325, 34151, 38286, 42742, 47531, 52665, 58156, 64016, 70257, 76891, 83930, 91386, 99271, 107597, 116376, 125620, 135341

Offset: 1

Author

Clark Kimberling, Jul 04 2012

Keywords

• nonn
• easy

Comments

Zhu Shijie gives in his Magnus Opus "Jade Mirror of the Four Unknowns" the problem: "The total number of apples in a pile in the form of a cone is 932, and the number of layers is an odd number." Zhu Shijie assumed the rational sequence s(k) = (k*(k+1)*(2*k+1)+k+1)/8 for the total number of apples in k layers, with n = (k+1)/2 is the solution 932 = a((15+1)/2) with k = 15. Zhu Shijie gave the solution polynomial: "Let the element tian be the number of layers. From the statement we have 7455 for the negative shi, 2 for the positive fang, 3 for the positive first lian, and 2 for the positive yu." This translates into the polynomial equation: 2*x^3 + 3*x^2 + 2*x - 7455 = 0. - Thomas Scheuerle, Feb 10 2025

References

• Zhu Shijie, Jade Mirror of the Four Unknowns (Siyuan yujian), Book III Guo Duo Die Gang (Piles of Fruit), Problem number 7, (1303).

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000
• Zhu Shijie, Jade Mirror of the Four Unknowns 2, Translation by Library of Chinese classics, original from 1303.
• Wikipedia, Jade Mirror of the Four Unknowns.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000326, A002411, A085473, A213771, A220084 (for a list of numbers of the form n*P(k,n) - (n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number).

Cf. A260260 (comment). [Bruno Berselli, Jul 22 2015]

Programs

Mathematica

(See A213771.)
LinearRecurrence[{4,-6,4,-1},{1,11,42,106},70] (* Harvey P. Dale, Mar 29 2025 *)

PARI

a(n) = (4*n^3-3*n^2+n)/2; \\ Altug Alkan, Dec 16 2017

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

G.f.: x*(1 + 7*x + 4*x^2)/(1 - x)^4.

a(n) = (4*n^2 - 3*n + 1)*n/2 = n*A002411(n) - (n-1)*A002411(n-1). - Bruno Berselli, Dec 11 2012

a(n) = n*A000326(n) + Sum_{i=0..n-1} A000326(i). - Bruno Berselli, Dec 18 2013

a(n) - a(n-1) = A085473(n-1). - R. J. Mathar, Mar 02 2025

E.g.f.: exp(x)*x*(1 + 4*x)*(2 + x)/2. - Elmo R. Oliveira, Aug 08 2025