A007283 -id:A007283 - cp's OEIS frontend

A110286 a(n) = 15*2^n.

15, 30, 60, 120, 240, 480, 960, 1920, 3840, 7680, 15360, 30720, 61440, 122880, 245760, 491520, 983040, 1966080, 3932160, 7864320, 15728640, 31457280, 62914560, 125829120, 251658240, 503316480, 1006632960, 2013265920, 4026531840, 8053063680, 16106127360

Offset: 0

Alexandre Wajnberg, Sep 07 2005

The first differences are the sequence itself. Doubling the terms gives the same sequence (beginning one step further).

Vincenzo Librandi, Table of n, a(n) for n = 0..235
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A007283, A020714, A005009, A005010, A005015, A005029.

Cf. A000079, A051916, A173787.

[15*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

15*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,15,30] (* Harvey P. Dale, Oct 19 2014 *)

a(n)=15<Charles R Greathouse IV, Oct 07 2015

G.f.: 15/(1-2x). - Philippe Deléham, Nov 23 2008
a(n) = A000079(n)*15 = A007283(n)*5 = A020714(n)*3. - Omar E. Pol, Dec 17 2008
a(n) = A173787(n+4,n). - Reinhard Zumkeller, Feb 28 2010
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
a(n) = 2*a(n-1) (with a(0)=15). - Vincenzo Librandi, Dec 26 2010
E.g.f.: 15*exp(2*x). - Stefano Spezia, May 15 2021

Edited by Omar E. Pol, Dec 16 2008

A030067 The "Semi-Fibonacci sequence": a(1) = 1; a(n) = a(n/2) (n even); a(n) = a(n-1) + a(n-2) (n odd).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 1, 6, 3, 9, 2, 11, 5, 16, 1, 17, 6, 23, 3, 26, 9, 35, 2, 37, 11, 48, 5, 53, 16, 69, 1, 70, 17, 87, 6, 93, 23, 116, 3, 119, 26, 145, 9, 154, 35, 189, 2, 191, 37, 228, 11, 239, 48, 287, 5, 292, 53, 345, 16, 361, 69, 430, 1, 431, 70, 501, 17, 518, 87, 605, 6, 611, 93

Offset: 1

David W. Wilson

This is the "semi-Fibonacci sequence". The distinct numbers that appear are called "semi-Fibonacci numbers", and are given in A030068.
a(2n+1) >= a(2n-1) + 1 is monotonically increasing. a(2n)/n can be arbitrarily small, as a(2^n) = 1. There are probably an infinite number of primes in the sequence. - Jonathan Vos Post, Mar 28 2006
From Robert G. Wilson v, Jan 17 2014: (Start)
Positions where k occurs:
   k: sequence
   -:-----------------------------
   1: A000079;
   2: 3*A000079 = A007283;
   3: 5*A000079 = A020714;
   4: none in the first 10^6 terms;
   5: 7*A000079 = A005009;
   6: 9*A000079 = A005010;
   7: none in the first 10^6 terms;
   8: none in the first 10^6 terms;
   9: 11*A000079 = A005015;
  10: none in the first 10^6 terms;
  11: 13*A000079 = A005029;
  12: none in the first 10^6 terms;
(End)
Any integer N which occurs in this sequence first occurs as an odd-indexed term a(2k-1) = A030068(k-1), and thereafter at indices (2k-1)*2^j, j=1,2,3,... (Both of these statements follow immediately from the definition of even-indexed terms.) No N can occur a second time as an odd-indexed term: This follows from the definition of these terms, a(2n+1) = a(2n) + a(2n-1) = a(2n-1) + a(n), which shows that the subsequence of odd-indexed terms (A030068) is strictly increasing, and therefore equal to the range (or: set) of the semi-Fibonacci numbers. - M. F. Hasler, Mar 24 2017
The lines in the logarithmic scatterplot of the sequence corresponds to sets of indices with the same 2-adic valuation. - Rémy Sigrist, Nov 27 2017
Define the partition subsum polynomial of an integer partition m of n where m = (m_1, m_2, ...m_k) by ps(m,x) = Product_{i=1..k} (1+x^m_i). Expanding ps(m,x) gives 1+a_1 x+a_2 x^2+...+a_n x^n, where a_j is the number of ways to form the subsum j from the parts of m. Then the number of partitions m of n for which ps(m,x) has no repeated root is a(n). - George Beck, Nov 07 2018

			a(1) = 1 by definition.
a(2) = a(1) = 1.
a(3) = 1 + 1 = 2.
a(4) = a(2) = 1.
a(5) = 2 + 1 = 3.
a(6) = a(3) = 2.
a(7) = 3 + 2 = 5.
a(8) = a(4) = 1.
a(9) = 5 + 1 = 6.
a(10) = a(5) = 3.

T. D. Noe, Table of n, a(n) for n = 1..10000
Abdulaziz M. Alanazi, Augustine O. Munagi and Darlison Nyirenda, Power Partitions and Semi-m-Fibonacci Partitions, arXiv:1910.09482 [math.CO], 2019.
George E. Andrews, Binary and Semi-Fibonacci Partitions, Journal of Ramanujan Society of Mathematics and Mathematics Sciences, honoring A.K. Agarwal's 70th birthday, 7:1(2019), 01-06.
Cristina Ballantine and George Beck, Partitions enumerated by self-similar sequences, arXiv:2303.11493 [math.CO], 2023.
George Beck, Semi-Fibonacci Partitions
Rémy Sigrist, Colored logarithmic scatterplot of the first 10000 terms (where the color is function of the 2-adic valuation of n)

Cf. A000045, A030068, A074364, A129092, A129100.

See A109671 for a variant.

import Data.List (transpose)
a030067 n = a030067_list !! (n-1)
a030067_list = concat $ transpose [scanl (+) 1 a030067_list, a030067_list]
-- Reinhard Zumkeller, Jul 21 2013, Jul 07 2013

f:=proc(n) option remember; if n=1 then RETURN(1) elif n mod 2 = 0 then RETURN(f(n/2)) else RETURN(f(n-1)+f(n-2)); fi; end;

semiFibo[1] = 1; semiFibo[n_?EvenQ] := semiFibo[n] = semiFibo[n/2]; semiFibo[n_?OddQ] := semiFibo[n] = semiFibo[n - 1] + semiFibo[n - 2]; Table[semiFibo[n], {n, 80}] (* Jean-François Alcover, Aug 19 2013 *)

a(n) = if(n==1, 1, if(n%2 == 0, a(n/2), a(n-1) + a(n-2)));
vector(100, n, a(n)) \\ Altug Alkan, Oct 12 2015

a=[1]; [a.append(a[-2]+a[-1] if n%2 else a[n//2-1]) for n in range(2, 75)]
print(a) # Michael S. Branicky, Jul 07 2022

Theorem: a(2n+1) - a(2n-1) = a(n). Proof: a(2n+1) - a(2n-1) = a(2n) + a(2n-1) - a(2n-2) - a(2n-3) = a(n) - a(n-1) + a(n-1) (induction) = a(n). - N. J. A. Sloane, May 02 2010
a(2^n - 1) = A129092(n) for n >= 1, where A129092 forms the row sums and column 0 of triangle A129100, which is defined by the nice property that column 0 of matrix power A129100^(2^k) = column k of A129100 for k > 0. - Paul D. Hanna, Dec 03 2008
G.f. g(x) satisfies (1-x^2) g(x) = (1+x-x^2) g(x^2) + x. - Robert Israel, Mar 23 2017

A079583 a(n) = 3*2^n - n - 2.

Original entry on oeis.org

1, 3, 8, 19, 42, 89, 184, 375, 758, 1525, 3060, 6131, 12274, 24561, 49136, 98287, 196590, 393197, 786412, 1572843, 3145706, 6291433, 12582888, 25165799, 50331622, 100663269, 201326564, 402653155, 805306338, 1610612705, 3221225440

Offset: 0

Benoit Cloitre, Jan 25 2003

Row sums of A132110. - Gary W. Adamson, Aug 09 2007
Consider the infinite sequence of strings x(1) = a, x(2) = aba, x(3) = ababbaba, ..., where x(n+1) = x(n).b^{n+1}.x(n), for n >= 1. Each x(n), for n >= 2, has borders x(1), x(2), ..., x(n-1), none of which cover x(n). The length of x(n+1) is 3*2^n-n-2. - William F. Smyth, Feb 29 2012
Number of edges in the rooted tree g[n] (n>=0) defined recursively in the following manner: denoting by P[n] the path on n vertices, we define g[0] =P[2] while g[n] (n>=1) is the tree obtained by identifying the roots of 2 copies of g[n-1] and one of the end-vertices of P[n+1]; the root of g[n] is defined to be the other end-vertex of P[n+1]. Roughly speaking, g[4], for example, is obtained from the planted full binary tree of height 5 by replacing the edges at the levels 1,2,3,4 with paths of lengths 4, 3, 2, and 1, respectively. - Emeric Deutsch, Aug 08 2013

T. Flouri, C. S. Iliopoulos, T. Kociumaka, S. P. Pissis, S. J. Puglisi, W. F. Smyth, W. Tyczynski, New and efficient approaches to the quasiperiodic characterization of a string, Proc. Prague Stringology Conf., 2012, 75-88.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tomas Flouri, Costas S. Iliopoulos, Solon P. Pissis, W. F. Smyth, On approximate string covering (draft, 2012).
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000295, A132110, A227712, A083329 (first differences).

I:=[1, 3, 8]; [n le 3 select I[n] else 4*Self(n-1)-5*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 23 2012

lst={};Do[AppendTo[lst, 3*2^n-n-2], {n, 0, 4!}];lst (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)
LinearRecurrence[{4,-5,2},{1,3,8},40] (* Vincenzo Librandi, Jun 23 2012 *)

a(n)=3<Charles R Greathouse IV, Feb 29 2012

a(0)=1, a(n) = 2*a(n-1) + n;
Binomial transform of [1, 2, 3, 3, 3, ...]. - Gary W. Adamson, Aug 09 2007
G.f.: (x^2-x+1)/((1-2*x)*(1-x)^2) = 3*U(0)x, where U(k) = 1 - (k+2)/(3*2^k - 18*x*4^k/(6*x*2^k - (k+2)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Jul 04 2012
a(n) = (A227712(n) - 1)/3 - Emeric Deutsch, Feb 18 2016
a(n) = A007283(n) - n - 2. - Miquel Cerda, Aug 07 2016
a(n) = A000225(n) + A000325(n). - Miquel Cerda, Aug 08 2016

A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 7, 13, 27, 53, 107, 213, 427, 853, 1707, 3413, 6827, 13653, 27307, 54613, 109227, 218453, 436907, 873813, 1747627, 3495253, 6990507, 13981013, 27962027, 55924053, 111848107, 223696213, 447392427, 894784853, 1789569707, 3579139413, 7158278827, 14316557653

Offset: 0

Michael Somos, Jun 17 1999

Number of positive integers requiring exactly n signed bits in the modified non-adjacent form representation. - Ralf Stephan, Aug 02 2003
The n-th entry (n>1) of the sequence is equal to the 1,1-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
Pisano period lengths:  1, 1, 6, 2, 2, 6, 6, 2, 18, 2, 10, 6, 12, 6, 6, 2, 8, 18, 18, 2, ... - R. J. Mathar, Aug 10 2012
For n >= 1, a(n) is the number of ways to tile a strip of length n+2 with blue squares and blue and red dominos, with the restriction that the first two tiles must be the same color. - Guanji Chen and Greg Dresden, Jul 15 2024

			G.f. = 2 + 3*x + 7*x^2 + 13*x^3 + 27*x^4 + 53*x^5 + 107*x^6 + 213*x^7 + 427*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wieb Bosma, Signed bits and fast exponentiation, Journal de théorie des nombres de Bordeaux, Vol. 13, No. 1 (2001), pp. 27-41.
Karl Dilcher and Larry Ericksen, Continued fractions and Stern polynomials, Ramanujan Journal 45.3 (2018): 659-681. See Table 2.
Karl Dilcher and Hayley Tomkins, Square classes and divisibility properties of Stern polynomials, Integers, Vol. 18 (2018), Article #A29.
Petro Kosobutskyy, The Collatz problem as a reverse n->0 problem on a graph tree formed from theta*2^n Jacobsthal-type numbers, arXiv:2306.14635 [math.GM], 2023.
Petro Kosobutskyy and Dariia Rebot, Collatz conjecture 3n+/-1 as a Newton binomial problem, Comp. Des. Sys. Theor. Prac., Lviv Nat'l Polytech. Univ. (Ukraine 2023) Vol. 5, No. 1, 137-145. See p. 140.
Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
Sam Northshield, Stern's diatomic sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, ..., Amer. Math. Monthly, Vol. 117, No. 7 (2010), pp. 581-598.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A084214 (first differences).
Cf. A001045, A014551, A024493, A062510.
Cf. A007283, A007679, A078008, A081254.
Cf. A083322, A083944, A083581, A084170.
Cf. A102713, A127978, A130755, A139763.
Cf. A140360, A140966, A166249, A168642.
Cf. A280560, A290604, A340627.
Cf. A112387, A135318.

[(5*2^n+(-1)^n)/3: n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

LinearRecurrence[{1,2},{2,3},40] (* Harvey P. Dale, Dec 11 2017 *)

{a(n) = if( n<0, 0, (5*2^n + (-1)^n) / 3)};

{a(n) = if (n<0 ,0, if( n<2, n+2, a(n-1) + 2*a(n-2)))};

[(5*2^n+(-1)^n)/3 for n in range(35)] # G. C. Greubel, Apr 10 2019

G.f.: (2 + x) / (1 - x - 2*x^2).
a(n) = (5*2^n + (-1)^n) / 3.
a(n) = 2^(n+1) - A001045(n).
a(n) = A084170(n)+1 = abs(A083581(n)-3) = A081254(n+1) - A081254(n) = A084214(n+2)/2.
a(n) = 2*A001045(n+1) + A001045(n) (note that 2 is the limit of A001045(n+1)/A001045(n)). - Paul Barry, Sep 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=-charpoly(A,-1). - Milan Janjic, Jan 27 2010
Equivalently, with different offset, a(n) = b(n+1) with b(0)=1 and b(n) = Sum_{i=0..n-1} (-1)^i (1 + (-1)^i b(i)). - Olivier Gérard, Jul 30 2012
a(n) = A000975(n-2)*10 + 5 + 2*(-1)^(n-2), a(0)=2, a(1)=3. - Yuchun Ji, Mar 18 2019
a(n+1) = Sum_{i=0..n} a(i) + 1 + (1-(-1)^n)/2, a(0)=2. - Yuchun Ji, Apr 10 2019
a(n) = 2^n + J(n+1) = J(n+2) + J(n+1) - J(n), where J is A001045. - Yuchun Ji, Apr 10 2019
a(n) = A001045(n+2) + A078008(n) = A062510(n+1) - A078008(n+1) = (A001045(n+2) + A062510(n+1))/2 = A014551(n) + 2*A001045(n). - Paul Curtz, Jul 14 2021
From Thomas Scheuerle, Jul 14 2021: (Start)
a(n) = A083322(n) + A024493(n).
a(n) = A127978(n) - A102713(n).
a(n) = A130755(n) - A166249(n).
a(n) = A007679(n) + A139763(n).
a(n) = A168642(n) XOR A007283(n).
a(n) = A290604(n) + A083944(n). (End)
From Paul Curtz, Jul 21 2021: (Start)
a(n) = 5*A001045(n) - A280560(n+1) = abs(A140360(n+1)) - A280560(n+1).
a(n) = 2^n + A001045(n+1) = A001045(n+3) - A000079(n).
a(n) = A001045(n+4) - A340627(n). (End)
a(n) = A001045(n+5) - A005010(n).
a(n+1) + a(n) = a(n+2) - a(n) = 5*2^n. - Michael Somos, Feb 22 2023
a(n) = A135318(2*n) + A135318(2*n+1) = A112387(2*n) + A112387(2*n+1). - Paul Curtz, Jun 26 2024
E.g.f.: (cosh(x) + 5*cosh(2*x) - sinh(x) + 5*sinh(2*x))/3. - Stefano Spezia, May 18 2025

Formula of Milan Janjic moved here from wrong sequence by Paul D. Hanna, May 29 2010

A118416 Triangle read by rows: T(n,k) = (2k-1)2^(n-1), 0 < k <= n.

Original entry on oeis.org

1, 2, 6, 4, 12, 20, 8, 24, 40, 56, 16, 48, 80, 112, 144, 32, 96, 160, 224, 288, 352, 64, 192, 320, 448, 576, 704, 832, 128, 384, 640, 896, 1152, 1408, 1664, 1920, 256, 768, 1280, 1792, 2304, 2816, 3328, 3840, 4352, 512, 1536, 2560, 3584, 4608, 5632, 6656, 7680

Offset: 1

Reinhard Zumkeller, Apr 27 2006

Row sums give A014477: Sum_{k=1..n} T(n,k) = A014477(n-1);
central terms give A118415; T(2*k-1,k) = A058962(k-1);
T(n,1) = A000079(n-1);
T(n,2) = A007283(n-1) for n > 1;
T(n,3) = A020714(n-1) for n > 2;
T(n,4) = A005009(n-1) for n > 3;
T(n,5) = A005010(n-1) for n > 4;
T(n,n-1) = A118417(n-1) for n > 1;
T(n,n) = A014480(n-1) = A118413(n,n);
A001511(T(n,k)) = A002024(n,k);
A003602(T(n,k)) = A002260(n,k).
The alternating row sums, Sum_{k=1..n} (-1)^(k+1)*T(n,k), are: (a) in odd rows, the central term, T(n,(n+1)/2) = A058962((n-1)/2); (b) in even rows, the negation of the average of the two central terms, -(T(2n,n) + T(2n,+1))/2 = -A018215(m/2). The absolute values of the alternating row sums give the plain row means, Sum_{k=1..n} T(n,k)/n; the alternating sign row means are (-2)^(n-1). - Gregory Gerard Wojnar, Feb 10 2024

			Triangle begins:
   1;
   2,   6;
   4,  12,  20;
   8,  24,  40,  56;
  16,  48,  80, 112, 144;
  32,  96, 160, 224, 288, 352;
  64, 192, 320, 448, 576, 704, 832;

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened

Cf. A118413, A117303, A054582.

a118416 n k = a118416_tabl !! (n-1) !! (k-1)
a118416_row 1 = [1]
a118416_row n = (map (* 2) $ a118416_row (n-1)) ++ [a014480 (n-1)]
a118416_tabl = map a118416_row [1..]
-- Reinhard Zumkeller, Jan 22 2012

A118416 := proc(n,k) 2^(n-1)*(2*k-1) ; end proc: # R. J. Mathar, Sep 04 2011

Flatten[Table[(2k-1)2^(n-1),{n,10},{k,n}]] (* Harvey P. Dale, Aug 26 2014 *)

from math import isqrt
def A118416(n): return (a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(1-a)+(n<<1)-1<Chai Wah Wu, Jun 20 2025

T(n,k) = 2*T(n-1,k), 1 <= k < n; T(n,n) = A014480(n-1).

A072405 Triangle T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 11, 14, 11, 5, 1, 1, 6, 16, 25, 25, 16, 6, 1, 1, 7, 22, 41, 50, 41, 22, 7, 1, 1, 8, 29, 63, 91, 91, 63, 29, 8, 1, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1, 1, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 1, 1, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11, 1

Offset: 0

Henry Bottomley, Jun 16 2002

Starting 1,0,1,1,1,... this is the Riordan array ((1-x+x^2)/(1-x), x/(1-x)). Its diagonal sums are A006355. Its inverse is A106509. - Paul Barry, May 04 2005

			Rows start as:
  1;
  1, 1;
  1, 1,  1; (key row for starting the recurrence)
  1, 2,  2,  1;
  1, 3,  4,  3,  1;
  1, 4,  7,  7,  4, 1;
  1, 5, 11, 14, 11, 5, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Row sums give essentially A003945, A007283, or A042950.
Cf. A072406 for number of odd terms in each row.
Cf. A051597, A096646, A122218 (identical for n > 1).
Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), A173119 (q=3), A173120 (q=-4).

T:= func< n,k | n lt 3 select 1 else Binomial(n,k) - Binomial(n-2,k-1) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 28 2021

t[2, 1] = 1; t[n_, n_] = t[, 0] = 1; t[n, k_] := t[n, k] = t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013, after Ralf Stephan *)

A072405(n, k) = if(n>2, binomial(n, k)-binomial(n-2, k-1), 1) \\ M. F. Hasler, Jan 06 2024

def T(n,k): return 1 if n<3 else binomial(n,k) - binomial(n-2,k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021

T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise.
T(n, k) = T(n-1, k-1) + T(n-1, k) starting with T(2, 0) = T(2, 1) = T(2, 2) = 1 and T(n, 0) = T(n, n) = 1.
G.f.: (1-x^2*y) / (1 - x*(1+y)). - Ralf Stephan, Jan 31 2005
From G. C. Greubel, Apr 28 2021: (Start)
Sum_{k=0..n} T(n, k) = (n+1)*[n<3] + 3*2^(n-2)*[n>=3].
T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = -1. (End)

A093562 (5,1) Pascal triangle.

Original entry on oeis.org

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170

Offset: 0

Wolfdieter Lang, Apr 22 2004

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+4*z)/(1-(1+x)*z).
The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.
Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.
A007318, A093563 (d=6), A228196, A228576.

a093562 n k = a093562_tabl !! n !! k
a093562_row n = a093562_tabl !! n
a093562_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302

Offset: 0

Wolfdieter Lang, Jul 16 2004

This is the fifth member, q=5, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.

David A. Corneth, Table of n, a(n) for n = 0..9999
Wolfdieter Lang, First 10 rows.

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012

a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; a(n, 0)=1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

Original entry on oeis.org

3, 8, 20, 16, 10, 28, 36, 18, 48, 14, 30, 76, 40, 88, 32, 108, 58, 60, 136, 70, 148, 78, 168, 44, 196, 50, 208, 72, 108, 76, 256, 130, 276, 46, 148, 50, 316, 328, 336, 348, 178, 90, 190, 388, 396, 22, 42, 448, 456, 114, 52, 238, 240, 250, 516, 176, 268, 270, 556

Offset: 1

Louis Mello (mellols(AT)aol.com), Mar 26 2001

nonn

Assuming Wall's conjecture (which is still open) allows one to calculate A001175(m) when m is a prime power since for any k >= 1: A001175(prime(n)^k) = a(n)*prime(n)^(k-1). For example: A001175(2^k) = 3*2^(k-1) = A007283(k-1).

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014), # 14.8.5.
A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, (2004).
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Wikipedia, Pisano period.

Cf. A001175, A000961, A071774, A003147, A296240.

a:= proc(n) option remember; local F, k, p;
      F:=[1,1]; p:=ithprime(n);
      for k while F<>[0,1] do
        F:=[F[2], irem(F[1]+F[2],p)]
      od: k
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 16 2015

Table[p=Prime[n]; a={1,0}; a0=a; k=0; While[k++; s=Mod[Plus@@a,p];a=RotateLeft[a]; a[[2]]=s; a!=a0]; k, {n,100}] (* T. D. Noe, Jun 12 2006 *)

for(n=1,100,s=1; while(sum(i=n,n+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))+sum(i=n+1,n+1+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))>0,s++); print1(s,","))

from itertools import count
from sympy import prime
def A060305(n):
    x, p = (1,1), prime(n)
    for k in count(1):
        if x == (0,1):
            return k
        x = (x[1], (x[0]+x[1]) % p) # Chai Wah Wu, May 31 2022

a(n) = A001175(prime(n)). - Jonathan Sondow, Dec 09 2017

a(n) = (3 - L(p))/2 * (p - L(p)) / A296240(n) for n >= 4, where p = prime(n) and L(p) = Legendre(p|5); so a(n) <= p-1 if p == +- 1 mod 5, and a(n) <= 2*p+2 if p == +- 2 mod 5. See Wall's Theorems 6 and 7. - Jonathan Sondow, Dec 10 2017

Corrected by Benoit Cloitre, Jun 04 2002

Name clarified by Jonathan Sondow, Dec 09 2017

A070812 a(n) = phi(gpf(n)) - gpf(phi(n)) = A000010(A006530(n)) - A006530(A000010(n)).

Original entry on oeis.org

0, -1, 2, 0, 3, -1, -1, 2, 5, 0, 9, 3, 2, -1, 14, -1, 15, 2, 3, 5, 11, 0, -1, 9, -1, 3, 21, 2, 25, -1, 5, 14, 3, -1, 33, 15, 9, 2, 35, 3, 35, 5, 1, 11, 23, 0, -1, -1, 14, 9, 39, -1, 5, 3, 15, 21, 29, 2, 55, 25, 3, -1, 9, 5, 55, 14, 11, 3, 63, -1, 69, 33, -1, 15, 5, 9, 65, 2, -1, 35, 41, 3, 14, 35, 21, 5, 77, 1, 9, 11, 25, 23, 15, 0, 93, -1, 5

Offset: 3

Labos Elemer, May 09 2002

sign

Value of commutator[A000010, A006530] at n.

			Cases of n when a(n) = 1, -1, 2 or 0 are listed in A070002, A070003, A070004, A007283 respectively. Further regular solutions: if a(n)=3, then n=7k, where k has prime divisors < 7; if a(n)=5, then n=11k, where k has no prime divisors >=11; if a(n)=25, then mostly (not always!) n=31k ...

Michael De Vlieger, Table of n, a(n) for n = 3..10000
Michael De Vlieger, Log log scatterplot of a(n)+2, n = 3..10^5.

Cf. A000010, A006530, A070777, A068211, A070002, A070003, A070004, A007283.

pf[x_] := Part[Reverse[Flatten[FactorInteger[x]]], 2] Table[EulerPhi[pf[u]]-pf[EulerPhi[u]], {u, 3, 128}]

gpf(n)=my(f=factor(n)[,1]);f[#f]
a(n)=gpf(n)-gpf(eulerphi(n))-1 \\ Charles R Greathouse IV, Feb 19 2013

a(n) = A070777(n) - A068211(n).

cp's OEIS Frontend

A110286 a(n) = 15*2^n.

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A030067 The "Semi-Fibonacci sequence": a(1) = 1; a(n) = a(n/2) (n even); a(n) = a(n-1) + a(n-2) (n odd).

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A079583 a(n) = 3*2^n - n - 2.

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A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

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A118416 Triangle read by rows: T(n,k) = (2*k-1)*2^(n-1), 0 < k <= n.

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A072405 Triangle T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise, read by rows.

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A118416 Triangle read by rows: T(n,k) = (2k-1)2^(n-1), 0 < k <= n.