A038792 -id:A038792 - cp's OEIS frontend

A026532 Ratios of successive terms are 3, 2, 3, 2, 3, 2, 3, 2, ...

1, 3, 6, 18, 36, 108, 216, 648, 1296, 3888, 7776, 23328, 46656, 139968, 279936, 839808, 1679616, 5038848, 10077696, 30233088, 60466176, 181398528, 362797056, 1088391168, 2176782336, 6530347008, 13060694016, 39182082048, 78364164096, 235092492288, 470184984576

Offset: 1

Clark Kimberling

Preface the series with a 1: (1, 1, 3, 6, 18, 36, ...); then the next term in the series = (1, 1, 3, 6, ...) dot (1, 2, 1, 2, ...). Example: 36 = (1, 1, 3, 6, 18) dot (1, 2, 1, 2, 1) = (1 + 2 + 3 + 12 + 18). - Gary W. Adamson, Apr 18 2009

Partial products of A176059. - Reinhard Zumkeller, Apr 04 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..700
Sean A. Irvine, Walks on Graphs.
José L. Ramírez, Bi-periodic incomplete Fibonacci sequences, Annales Mathematicae et Informaticae 42 (2013), 83-92. See the 1st column of Table 1 on p. 85.
Index entries for linear recurrences with constant coefficients, signature (0,6).

Cf. A026519, A026520, A026521, A026522, A026523, A026524, A026525, A026526, A026527, A026528, A026529, A026530, A026531, A026533, A026534, A027262, A027263, A027264, A027265, A027266.
Cf. A026534, A026549, A176059, A208131.
Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

a026532 n = a026532_list !! (n-1)
a026532_list = scanl (*) 1 $ a176059_list
-- Reinhard Zumkeller, Apr 04 2012

[(1/4)*(3-(-1)^n)*6^Floor(n/2) : n in [1..30]]; // Vincenzo Librandi, Jun 08 2011

FoldList[(2 + Boole[EvenQ@ #2]) #1 &, Range@ 28] (* or *)
CoefficientList[Series[x*(1+3x)/(1-6x^2), {x,0,31}], x] (* Michael De Vlieger, Aug 02 2017 *)
LinearRecurrence[{0,6},{1,3},30] (* Harvey P. Dale, Jul 11 2018 *)

a(n)=if(n%2,3,1)*6^(n\2) \\ Charles R Greathouse IV, Jul 02 2013

def a(n): return (3 if n%2 else 1)*6**(n//2)
print([a(n) for n in range(31)]) # Indranil Ghosh, Aug 02 2017

[(1/2)*6^((n-2)/2)*(3*(1+(-1)^n) + sqrt(6)*(1-(-1)^n)) for n in (1..30)] # G. C. Greubel, Dec 21 2021

a(n) = T(n, 0) + T(n, 1) + ... + T(n, 2n-2), T given by A026519.
From Benoit Cloitre, Nov 14 2003: (Start)
a(n) = (1/2)*(5+(-1)^n)*a(n-1) for n>1, a(1) = 1.
a(n) = (1/4)*(3-(-1)^n)*6^floor(n/2).  (End)
From Ralf Stephan, Feb 03 2004: (Start)
G.f.: x*(1+3*x)/(1-6*x^2).
a(n+2) = 6*a(n). (End)
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = (1/2)*6^((n-2)/2)*(3*(1+(-1)^n) + sqrt(6)*(1-(-1)^n)). - G. C. Greubel, Dec 21 2021
Sum_{n>=1} 1/a(n) = 8/5. - Amiram Eldar, Feb 13 2023

New definition from Ralf Stephan, Dec 01 2004

Offset changed from 0 to 1 by Vincenzo Librandi, Jun 08 2011

A134511 abs(A049310) * A128174 provided both arrays are read with offset (n,k) = (0,0).

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 3, 0, 1, 5, 0, 4, 0, 1, 0, 8, 0, 5, 0, 1, 13, 0, 12, 0, 6, 0, 1, 0, 21, 0, 17, 0, 7, 0, 1, 34, 0, 33, 0, 23, 0, 8, 0, 1, 0, 55, 0, 50, 0, 30, 0, 9, 0, 1, 89, 0, 88, 0, 73, 0, 38, 0, 10, 0, 1, 0, 144, 0, 138, 0, 103, 0, 47, 0, 11, 0, 1, 233, 0, 232, 0, 211, 0, 141, 0, 57, 0, 12, 0, 1

Offset: 0

Gary W. Adamson, Oct 28 2007

A112552(unsigned) = A128174 * A049310.
Row sums = A134512: (1, 1, 3, 4, 10, 14, 32, 46, 99, 145, ...).
From Petros Hadjicostas, Sep 03 2019: (Start)
To prove Alois P. Heinz's claim (see the Formula section and his Maple program below) we note that, for n >= 0 and 0 <= k <= n, T(n, n-k) = Sum_{r = 0 .. infinity} abs(A049310(n,r)) * A128174(r,n-k) = Sum_{r = n-k..n} abs(A049310(n,r)) * A128174(r,n-k). But A049310(n,r) = 0 when n + r is odd and A128174(r,n-k) = 1 iff r + n - k is even. Thus, when k is odd, T(n, n-k) = 0.
Assume now k is even. Then T(n, n-k) = Sum_{r = n-k..n and n+r even} abs(A049310(n,r)) =  Sum_{r = n-k..n and n+r even} binomial((n+r)/2, r). Letting m = n-r (which is even), we see that the summation ranges from m = 0 to k over even numbers. Thus, let s = m/2, and so T(n, n-k) = Sum_{s = 0 .. k/2} binomial(n-s, n-2*s) = Sum_{s = 0 .. k/2} binomial(n-s, s) = F(n+1, k/2), where F(.,.) is the incomplete Fibonacci number from the references (see also the Formula section below).
(End)

			First few rows of the triangle T(n,k):
   1;
   0,  1;
   2,  0,  1;
   0,  3,  0,  1;
   5,  0,  4,  0,  1;
   0,  8,  0,  5,  0,  1;
  13,  0, 12,  0,  6,  0,  1;
   0, 21,  0, 17,  0,  7,  0,  1;
  34,  0, 33,  0, 23,  0,  8,  0,  1;
   0, 55,  0, 50,  0, 30,  0,  9,  0,  1;
  ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 141, flattened)
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 (p. 39) that contains the incomplete Fibonacci numbers.
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A038730, A038792, A049310, A128174, A134512.

A(4n,2n) gives: A038736.

N:= 20: # for the first N rows
T128174:= Matrix(N,N,(i,j) -> `if`(j<=i, (i-j+1) mod 2, 0)):
T049310:= Matrix(N,N):
for i from 1 to N do
     P:= orthopoly[U](i-1,x/2);
     for j from 1 to i do
       T049310[i,j]:= abs(coeff(P,x,j-1))
     od
od:
A:= T049310 . T128174:
for i from 1 to N do
convert(A[i,1..i],list)
od;  # Robert Israel, Mar 02 2018
# second Maple program:
T:= (n, k)-> `if`((n+k)::odd, 0, add(binomial(n-s, s), s=0..(n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..12); # Alois P. Heinz, Sep 02 2019

T[n_, k_] := If[OddQ[n+k], 0, Sum[Binomial[n-s, s], {s, 0, (n-k)/2}]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Dec 31 2021, after Alois P. Heinz *)

abs(A049310) * A128174 as infinite lower triangular matrices assuming both of them have offset (n,k) = (0,0).
From Petros Hadjicostas, Sep 03 2019: (Start)
Let F(m,r) = Sum_{j = 0..r} binomial(m-1-j, j) be the incomplete Fibonacci numbers from the references (defined for m >= 1 and 0 <= r <= floor((m-1)/2)).
As Alois P. Heinz observed, for n >= 0 and 0 <= k <= n, T(n, n-k) = F(n+1, k/2) when k is even, and = 0 otherwise (see his Maple program below).
(End)

Edited by Robert Israel, Mar 02 2018

A038791 An intermediate sequence for nonisomorphic circulant p^2-tournaments, indexed by odd primes p.

Original entry on oeis.org

2, 4, 12, 104, 344, 4096, 14572, 190652, 9586984, 35791472, 1908874584, 27487790720, 104715393912, 1529755308212, 86607685141744, 4969489243995032, 19215358410149344, 1117984489315857512, 16865594581677305360, 65588423373189982912

Offset: 2

N. J. A. Sloane, May 04 2000

Number of subsets of {1, ..., p} with product = 1 mod p, where p is the n-th prime. - Charles R Greathouse IV, Jun 06 2013

Also : Number of subsets of {1, ..., p} with product = -1 mod p, where p is the n-th prime. - Ridouane Oudra, Jul 08 2025

Charles R Greathouse IV, Table of n, a(n) for n = 2..100
M. Klin, V. A. Liskovets and R. Poeschel, Analytical enumeration of circulant graphs with prime-squared vertices, Sem. Lotharingien de Combin., B36d, 1996, 36 pages.

Cf. A038787, A238446.

has[p_] := Module[{v, u}, v = Table[0, {p-1}]; v[[1]] = 1; For[n = 2, n <= p-1, n++, u = Table[0, {p-1}]; For[j = 1, j <= p-1, j++, u[[Mod[j*n, p]]] += v[[j]]]; v += u]; 2*v[[1]]];
a[n_] := has[Prime[n]];
Table[a[n], {n, 2, 21}] (* Jean-François Alcover, Aug 30 2019, after Charles R Greathouse IV *)

has(p)=my(v=vector(p-1),u); v[1]=1; for(n=2,p-1,u=vector(p-1); for(j=1,p-1, u[j*n%p]+=v[j]);v+=u); 2*v[1]
a(n)=has(prime(n)) \\ Charles R Greathouse IV, Jun 06 2013

a(p^2) = A038790(p^2) - A038789(p^2) + A038792(p^2).

a(n) = A238446(n) + 1. - Ridouane Oudra, Jul 08 2025

More terms from Valery A. Liskovets, May 09 2001

a(12)-a(20) from Charles R Greathouse IV, Jun 06 2013

A137176 Hyperlucas number array T(r,n) = L(n)^(r), read by ascending antidiagonals (r >= 0, n >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 4, 4, 0, 1, 5, 8, 7, 0, 1, 6, 13, 15, 11, 0, 1, 7, 19, 28, 26, 18, 0, 1, 8, 26, 47, 54, 44, 29, 0, 1, 9, 34, 73, 101, 98, 73, 47, 0, 1, 10, 43, 107, 174, 199, 171, 120, 76, 0, 1, 11, 53, 150, 281, 373, 370, 291, 196, 123

Offset: 0

Jonathan Vos Post, Apr 04 2008

In Theorem 17, Dil and Mezo (2008) connect the hyperlucas numbers (this array) with the incomplete Lucas numbers (A324242). - Petros Hadjicostas, Sep 03 2019

			The array T(r,n) = L(n)^(r) begins:
.....|n=0|n=1|.n=2|.n=3|.n=4.|.n=5.|..n=6.|.n=7..|..n=8..|..n=9..|.n=10..|.in.OEIS
r=0..|.0.|.1.|..3.|..4.|...7.|..11.|...18.|...29.|....47.|....76.|...123.|.A000204
r=1..|.0.|.1.|..4.|..8.|..15.|..26.|...44.|...73.|...120.|...196.|...319.|.A027961
r=2..|.0.|.1.|..5.|.13.|..28.|..54.|...98.|..171.|...291.|...487.|...806.|.A023537
r=3..|.0.|.1.|..6.|.19.|..47.|.101.|..199.|..370.|...661.|..1148.|..1954.|.A027963
r=4..|.0.|.1.|..7.|.26.|..73.|.174.|..373.|..743.|..1404.|..2552.|..4506.|.A027964
r=5..|.0.|.1.|..8.|.34.|.107.|.281.|..654.|.1397.|..2801.|..5353.|..9859.|.A053298
r=6..|.0.|.1.|..9.|.43.|.150.|.431.|.1085.|.2482.|..5283.|.10636.|.20495.|.new
r=7..|.0.|.1.|.10.|.53.|.203.|.634.|.1719.|.4201.|..9484.|.20120.|.40615.|.new
r=8..|.0.|.1.|.11.|.64.|.267.|.901.|.2620.|.6821.|.16305.|.36425.|.77040.|.new
r=9..|.0.|.1.|.12.|.76.|.343.|1244.|.3864.|10685.|.26990.|.63415.|140455.|.new
For example, T(4,5) = L(5)^(4) = L(0)^(3) + L(1)^(3) + L(2)^(3) + L(3)^(3) + L(4)^(3) + L(5)^(3) = 0 + 1 + 6 + 19 + 47 + 101 = 174. - _Petros Hadjicostas_, Sep 03 2019

Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, arXiv:0803.4388 [math.NT], 2008.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, Applied Mathematics and Computation 206(2) (2008), 942-951.
Eric W. Weisstein, Steenrod Algebra.

Cf. A000204, A027961, A023537, A027963, A027964, A053298, A123736, A136431.

Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

L:= proc(r, n) option remember; `if`(n=0, 0, `if`(r=0,
      `if`(n<3, 2*n-1, L(0, n-2)+L(0, n-1)), L(r-1, n)+L(r, n-1)))
    end:
seq(seq(L(d-n, n), n=0..d), d=0..12);  # Alois P. Heinz, Sep 03 2019

L[r_, n_] := L[r, n] = If[n == 0, 0, If[r == 0, If[n < 3, 2n-1, L[0, n-2] + L[0, n-1]], L[r-1, n] + L[r, n-1]]];
Table[L[d-n, n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

T(r,n) = L(n)^(r) = Apply partial sum operator r times to Lucas numbers A000204.
From Petros Hadjicostas, Sep 03 2019: (Start)
T(r, n) = L(n)^(r) = Sum_{k = 0..n} L(k)^(r-1) for r >= 1, with T(0,n) = L(n)^(0) = L(n) = A000204(n), T(r,0) = L(0)^(r) = 0, and T(r,1) = L(1)^(r) = 1. (See Definition 13 in Dil and Mezo (2008).)
G.f. for row r: Sum_{n >= 0} L(n)^(r)*t^n = t * (1+2*t)/((1-t-t^2) * (1-t)^r). (Corrected from Proposition 14 in Dil and Mezo (2008).)
(End)

A324242 Incomplete Lucas numbers: irregular triangular array L(n,k) = Sum_{j = 0..k} (n/(n-j)) * binomial(n-j, j), read by rows, with n >= 1 and 0 <= k <= floor(n/2).

Original entry on oeis.org

1, 1, 3, 1, 4, 1, 5, 7, 1, 6, 11, 1, 7, 16, 18, 1, 8, 22, 29, 1, 9, 29, 45, 47, 1, 10, 37, 67, 76, 1, 11, 46, 96, 121, 123, 1, 12, 56, 133, 188, 199, 1, 13, 67, 179, 284, 320, 322, 1, 14, 79, 235, 417, 508, 521, 1, 15, 92, 302, 596, 792, 841, 843, 1, 16, 106, 381, 831, 1209, 1349, 1364

Offset: 1

Petros Hadjicostas, Sep 02 2019

For additional properties of the incomplete Lucas numbers and special cases not listed here, see Filipponi (1996, pp. 45-53).

			Triangle L(n,k) (with rows n >= 1 and columns k >= 0) begins as follows:
  1;
  1,  3;
  1,  4;
  1,  5,  7;
  1,  6, 11;
  1,  7, 16,  18;
  1,  8, 22,  29;
  1,  9, 29,  45,  47;
  1, 10, 37,  67,  76;
  1, 11, 46,  96, 121, 123;
  1, 12, 56, 133, 188, 199;
  ...
Row sums are 1, 4, 5, 13, 18, 42, 60, 131, 191, 398, 589, 1186, 1775, 3482, 5257, 10103, 15360, ...

A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Lucas numbers.
Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 2 (p. 46) that contains the incomplete Lucas numbers.
A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Cf. A038730, A038792, and A134511 for various versions of the incomplete Fibonacci numbers.

Cf. A000032, A000045, A000124, A000204, A137176.

Flatten[Table[Sum[(n/(n-j))*Binomial[n-j, j],{j,0,k}],{n,1,15},{k,0,Floor[n/2]}]] (* Stefano Spezia, Sep 03 2019 *)

L(n,k) = F(n-1, k-1) + F(n+1, k) for n >= 1 and 0 <= k <= floor(n/2), where F(n,k) = Sum_{j = 0..k} binomial(n-1-j, j) are the incomplete Fibonacci numbers (defined for n >= 1 and 0 <= k <= floor((n-1)/2)).
L(n+2, k+1) = L(n+1, k+1) + L(n,k) for n >= 1 and 0 <= k <= floor((n-1)/2).
L(n,k) = F(n+2,k) - F(n-2, k-2) for n >= 3 and 2 <= k <= floor((n+1)/2).
Special cases: L(n,0) = 1 (n >= 1), L(n,1) = n+1 (n >= 2), L(n,2) = (n^2-n+2)/2 = A000124(n-1) (n >= 4), and L(n, floor(n/2)) = A000204(n) (n >= 1).
Sum of row n = (3 + (-1)^n)*A000204(n)/4 + n*A000045(n)/2.
G.f. for column k >= 1: t^(2*k)*((A000204(2*k) + t*A000204(2*k-1))*(1-t)^(k+1) - t^2*(2-t))/((1-t)^(k+1) * (1-t-t^2)).

cp's OEIS Frontend

A038799 T(2n+6,n), array T as in A038792.

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A026532 Ratios of successive terms are 3, 2, 3, 2, 3, 2, 3, 2, ...

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A134511 abs(A049310) * A128174 provided both arrays are read with offset (n,k) = (0,0).

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A038791 An intermediate sequence for nonisomorphic circulant p^2-tournaments, indexed by odd primes p.

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A137176 Hyperlucas number array T(r,n) = L(n)^(r), read by ascending antidiagonals (r >= 0, n >= 0).

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A324242 Incomplete Lucas numbers: irregular triangular array L(n,k) = Sum_{j = 0..k} (n/(n-j)) * binomial(n-j, j), read by rows, with n >= 1 and 0 <= k <= floor(n/2).

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