A079824 Sum of numbers in n-th upward diagonal of triangle in A079823.

1, 2, 7, 12, 25, 37, 62, 84, 125, 160, 221, 272, 357, 427, 540, 632, 777, 894, 1075, 1220, 1441, 1617, 1882, 2092, 2405, 2652, 3017, 3304, 3725, 4055, 4536, 4912, 5457, 5882, 6495, 6972, 7657, 8189, 8950, 9540, 10381, 11032, 11957, 12672, 13685, 14467, 15572

Offset: 1

Author

Amarnath Murthy, Feb 11 2003

Keywords

• base
• easy
• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Crossrefs

Cf. A000326, A079823, A185787 and A185788 (bisections).

Programs

Magma

[(15+25*n+15*n^2+14*n^3 -3*(-1)^n*(5+3*n+n^2))/96: n in [1..60]]; // G. C. Greubel, Dec 08 2023

Maple

A079824aux := proc(n,k)
    A000124(n)+k ;
end proc:
A079824 := proc(n)
    local a,k,n0 ;
    n0 := n-1 ;
    a := 0 ;
    for k from 0 to floor(n0/2) do
        a := a+A079824aux(n0-k,k) ;
    end do:
    a ;
end proc: # R. J. Mathar, Aug 23 2012

Mathematica

LinearRecurrence[{1,3,-3,-3,3,1,-1},{1,2,7,12,25,37,62},60] (* Harvey P. Dale, May 06 2014 *)

Python

def a(n): return (15 + 25*n + 15*(n**2) + 14*(n**3) - 3*(((-1)**n))*(5 + n*(3 + n))) // 96 # Torlach Rush, Aug 14 2022

SageMath

[(15+25*n+15*n^2+14*n^3 -3*(-1)^n*(5+3*n+n^2))/96 for n in range(1,61)] # G. C. Greubel, Dec 08 2023

Formula

From Philippe Deléham, Feb 16 2004: (Start)

a(2*n) = (n/6)*(7*n^2 + 3*n + 2);

a(2*n-1) = (n/6)*(7*n^2 - 6*n + 5). (End)

G.f.: x*(1+x+2*x^2+2*x^3+x^4) / ( (1+x)^3*(1-x)^4 ). - R. J. Mathar, Aug 23 2012

From Richard Peterson, Aug 19 2020: (Start)

a(2*n) - a(2*n-1) = A000326(n).

a(2*n+1) - a(2*n) = n^2 + (n+1)^2. (End)

a(n) = (15 + 25*n + 15*n^2 + 14*n^3 - 3*(-1)^n*(5 + n*(3 + n)))/96. - Torlach Rush, Aug 14 2022

E.g.f.: (1/48)*( x*(33 + 27*x + 7*x^2)*cosh(x) + (15 + 21*x + 30*x^2 + 7*x^3)*sinh(x) ). - G. C. Greubel, Dec 08 2023

Extensions

More terms from Jason D. W. Taff (jtaff(AT)jburroughs.org), Oct 31 2003

More terms from Philippe Deléham, Feb 16 2004

Typo corrected by Kevin Ryde, Aug 23 2012