A129696 -id:A129696 - cp's OEIS frontend

A227356 Partial sums of A129361.

1, 2, 5, 10, 20, 36, 65, 112, 193, 324, 544, 900, 1489, 2442, 4005, 6534, 10660, 17336, 28193, 45760, 74273, 120408, 195200, 316216, 512257, 829458, 1343077, 2174130, 3519412, 5696124, 9219105, 14919408, 24144289

Offset: 1

Kival Ngaokrajang, Jul 08 2013

nonn

Sum of labeled numbers of boxes arranged as Pyramid type-II with base Fibonacci(n).
Let us call a Pyramid "type-I" when each row starts with the same number as the leftmost base number, and "type-II" when each column has the same number as the base.
The Pyramid arrangements are related to other sequences as follows:
   Base Number     Type-I     Type-II
   -----------     ------     -------
   Natural         A002623    A034828
   Odd             A000292    A128624
   Fibonacci       A129696    a(n)
   1               A002620    A002620
   1,0             A008805
See illustration in links.

Kival Ngaokrajang, Illustration for some small n.
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,1,-1,0,1).

Cf. A002623, A034828, A002620, A000292, A128624, A129696, A008805.

LinearRecurrence[{2,1,-3,1,-1,0,1},{1,2,5,10,20,36,65},40] (* Harvey P. Dale, Jun 30 2025 *)

For n >=2, a(n) = a(n-1) + A129361(n-1).
G.f. -x*(1+x)*(x^2-x+1) / ( (x-1)*(x^2+x-1)*(x^4+x^2-1) ). - Joerg Arndt, Jul 10 2013
a(n) = 2 + A000045(n+4) - A096748(n+6). - R. J. Mathar, Jul 20 2013

A230449 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A052952(n), n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 5, 4, 1, 4, 8, 9, 8, 1, 5, 12, 17, 17, 12, 1, 6, 17, 29, 34, 29, 21, 1, 7, 23, 46, 63, 63, 50, 33, 1, 8, 30, 69, 109, 126, 113, 83, 55, 1, 9, 38, 99, 178, 235, 239, 196, 138, 88, 1, 10, 47, 137, 277, 413, 474, 435, 334, 226, 144

Offset: 0

Johannes W. Meijer, Oct 19 2013

The right hand columns of triangle T(n, k) represent the Kn2p sums of the ‘Races with Ties’ triangle A035317. See A180662 for the definitions of these sums.
The row sums lead to A094687, the convolution of Fibonacci and Jacobsthal numbers, and the alternating row sums lead to A008346.
The backwards antidiagonal sums equal Kn21(n) = (-1)^n*A175722(n).

			The first few rows of triangle T(n, k), n >= 0 and 0 <= k <= n.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1
1|  1,  1
2|  1,  2,  3
3|  1,  3,  5,   4
4|  1,  4,  8,   9,   8
5|  1,  5, 12,  17,  17,   12
6|  1,  6, 17,  29,  34,   29,   21
7|  1,  7, 23,  46,  63,   63,   50,   33
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1,  1,  3,   4,   8,   12,   21,   33
1|  1,  2,  5,   9,  17,   29,   50,   83
2|  1,  3,  8,  17,  34,   63,  113,  196
3|  1,  4, 12,  29,  63,  126,  239,  435
4|  1,  5, 17,  46, 109,  235,  474,  909
5|  1,  6, 23,  69, 178,  413,  887, 1796
6|  1,  7, 30,  99, 277,  690, 1577, 3373
7|  1,  8, 38, 137, 414, 1104, 2681, 6054

Cf. (Triangle columns) A000012, A000027, A089071, A052952, A129696

Cf. A230447, A230448

T:= proc(n, k) option remember: if k=0 then return(1) elif k=n then return(combinat[fibonacci](n+2) - (1-(-1)^n)/2) else procname(n-1,k-1)+procname(n-1,k) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # End first program.
T := proc(n, k): add(A035317(k-p+n-k, k-2*p), p=0..floor(k/2)) end: A035317 := proc(n, k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # End second program.

T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = F(n+2) - (1-(-1)^n)/2 = A052952(n), with F(n) = A000045(n), the Fibonacci numbers, n >= 0 and 0 <= k <= n.
T(n+p-1, n) = sum(A035317(n-k+p-1, n-2*k), k=0..floor(n/2)), n >= 0 and p >= 1.
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
Tsq(n, k) = sum(Tsq(n-1, i), i=0..k), n >= 1 and k >= 0, with Tsq(0, k) = A052952(k).
Tsq(n, k) = sum(A035317(n+k-i, k-2*i), i=0..floor(k/2)), n >= 0 and k >= 0.
Tsq(n, k) = A052952(2*n+k) - sum(A035317(n+k+i+1, k+2*i+2), i = 0..n-1)
The G.f. generates the terms in the n-th row of the square array Tsq(n, k).
G.f.: (-1)^(n)/((-1+x+x^2)*(x+1)*(x-1)^(n+1)), n >= 0.

A215006 a(0)=0, a(n+1) is the least k>a(n) such that k+a(n)+n+1 is a Fibonacci number.

Original entry on oeis.org

0, 1, 2, 3, 6, 10, 18, 30, 51, 84, 139, 227, 371, 603, 980, 1589, 2576, 4172, 6756, 10936, 17701, 28646, 46357, 75013, 121381, 196405, 317798, 514215, 832026, 1346254, 2178294, 3524562, 5702871, 9227448, 14930335, 24157799, 39088151, 63245967, 102334136, 165580121

Offset: 0

Alex Ratushnyak, Jul 31 2012

Same definition for k:
k+b(n)+n is a square for each term b(n) of A097063 except the first;
k+b(n)+n+1 is a square for each term b(n) of A007590 except the first;
k+b(n)+n is a cube for each term b(n) of the sequence 0, 7, 18, 43, 78, 133, 204, 301, 420, 571, 750, 967, 1218, 1513, 1848, 2233, 2664, 3151, 3690, 4291, 4950, 5677, 6468, 7333, ... (last digit repeats with period 10);
k+b(n)+n is a triangular number for each term b(n) of A002378 (oblong numbers);
k+b(n)+n is an oblong number for each term b(n) of A000217 (triangular numbers);
k+b(n)+n is a prime for each term b(n) of the sequence 0, 1, 2, 6, 7, 11, 12, 18, 21, 23, 26, 30, 31, 35, 40, 42, 43, 47, 48, 60, 69, 73, 78, 80, 87, 99, 102, 104, 107, 115, 118, 120, 125, 135, ...

			For n + 1 = 7, a(n + 1) = 30 is the least k > a(n) = a(6) = 18 such that k + a(n) + n + 1 = 30 + 18 + 6 + 1 = 55 is a Fibonacci number. - _David A. Corneth_, Sep 03 2016

Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A000045, A000217, A002378, A007590, A097063.

[n le 3 select n else Self(n)+Self(n-1)+Floor(n/2)-1: n in [0..40]]; // Bruno Berselli, Jul 31 2012

Join[{0}, LinearRecurrence[{2, 1, -3, 0, 1}, {1, 2, 3, 6, 10}, 39]] (* Jean-François Alcover, Oct 05 2017 *)

prpr = 0
prev = 1
fib = [0]*100
for n in range(100):
    fib[n] = prpr
    curr = prpr+prev
    prpr = prev
    prev = curr
a = 0
for n in range(1,55):
    print(a, end=',')
    b = c = 0
    while c <= a:
        c = fib[b] - a - n
        b += 1
    a=c

print(0, end=',')
prpr = 1
prev = 2
for n in range(3,56):
    print(prpr, end=',')
    curr = prpr+prev + n//2 - 1
    prpr = prev
    prev = curr

a(n) = a(n-1) +a(n-2) +floor(n/2) -1 with n>1, a(0)=0, a(1)=1.
From Bruno Berselli, Jul 31 2012: (Start)
G.f.: x*(1-2*x^2+x^3+x^4)/((1+x)*(1-x)^2*(1-x-x^2)).
a(n) = Fibonacci(n+2)-A004526(n+1) with n>0, a(0)=0.
a(n) = A129696(n-1)+1 with n>1, a(0)=0, a(1)=1. (End)

Definition corrected by David A. Corneth, Sep 03 2016

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A227356 Partial sums of A129361.

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A230449 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A052952(n), n >= 0 and 0 <= k <= n.

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A215006 a(0)=0, a(n+1) is the least k>a(n) such that k+a(n)+n+1 is a Fibonacci number.

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