A002307 -id:A002307 - cp's OEIS frontend

A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.

1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344, 302408598355968

Offset: 0

N. J. A. Sloane, Mira Bernstein, Simon Plouffe

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths:  1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and  B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) =  1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
A. Bernini, F. Disanto, R. Pinzani, and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7.
A. Burstein, S. Kitaev, and T. Mansour, Independent sets in certain classes of (almost) regular graphs, arXiv:math/0310379 [math.CO], 2003.
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607. (See Cor. 3.7(e).)
Tomislav Doslic and I. Zubac, Counting maximal matchings in linear polymers, Ars Mathematica Contemporanea 11 (2016) 255-276.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
L. Escobar, P. Gallardo, J. González-Anaya, J. L. González, G. Montúfar, and A. H. Morales, Enumeration of max-pooling responses with generalized permutohedra, arXiv:2209.14978 [math.CO], 2022. (See Ex. 5.1)
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Pamela Fleischmann, Jonas Höfer, Annika Huch, and Dirk Nowotka, alpha-beta-Factorization and the Binary Case of Simon's Congruence, arXiv:2306.14192 [math.CO], 2023.
A. S. Fraenkel and C. Kimberling, Generalized Wythoff arrays, shuffles and interspersions, Discr. Math. 126 (1-3) (1994) 137-149.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Sela Fried, Toufik Mansour, and Mark Shattuck, Counting k-ary words by number of adjacency differences of a prescribed size, arXiv:2504.03013 [math.CO], 2025. See p. 7.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 440
László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222. [Annotated scanned copy]
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.
Index entries for sequences related to poker
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Row sums of A059474. - David W. Wilson, Aug 14 2006
Cf. A000129, A000225, A002307, A002315, A006012 (same recurrence), A007052, A075271, A078057, A106731, A179260, A204089, A228405.
Equals 2 * A003480, n>0.
Row sums of A140071.
Cf. A127672, A265185, A133314.

a007070 n = a007070_list !! n
a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
   (tail $ map (* 2) a007070_list) a007070_list)
-- Reinhard Zumkeller, Jan 16 2012

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011

[n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011

A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015

LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)

a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)

a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))

[lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x  where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024

A097159 Smallest prime p such that there are n consecutive quadratic residues mod p.

Original entry on oeis.org

2, 7, 11, 19, 43, 67, 83, 131, 283, 277, 467, 479, 1907, 1607, 2543, 1559, 5443, 5711, 6389, 14969, 25703, 10559, 20747, 52057, 136223, 90313, 162263, 18191, 167107, 31391, 376589, 607153, 671947

Offset: 1

Robert G. Wilson v, Jul 28 2004

nonn

Additional terms less than 10^6: a(35)=298483, a(36)=422231, a(40)=701399 and a(42)=366791. - T. D. Noe, Apr 03 2007

			a(22)=10559, a(23)=20747 & a(28)=18191.

Cf. A002307, A097160, A097161.

Cf. A129201.

f[l_, a_] := Module[{A = Split[l], B}, B = Last[ Sort[ Cases[A, x : {a ..} :> { Length[x], Position[A, x][[1, 1]] }] ]]; {First[B], Length[ Flatten[ Take[A, Last[B] - 1]]] + 1}]; g[n_] := g[n] = f[ JacobiSymbol[ Range[ Prime[n] - 1], Prime[n]], 1][[1]]; g[1] = 1; a = Table[0, {30}]; Do[b = g[n]; If[ a[[b]] < 31 && a[[b]] == 0, a[[b]] = n; Print[b, " = ", Prime[n]]], {n, 2555}]

More terms from T. D. Noe, Apr 03 2007

A002308 Consecutive quadratic nonresidues mod p.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 3, 4, 4, 3, 4, 4, 5, 5, 4, 6, 5, 6, 6, 6, 4, 6, 7, 6, 6, 5, 7, 6, 10, 4, 7, 8, 5, 5, 6, 7, 5, 6, 6, 5, 6, 6, 6, 5, 5, 6, 7, 7, 7, 6, 7, 6, 5, 7, 6, 7, 9, 7, 7, 7, 9, 5, 7, 10, 7, 7, 8, 7, 8, 6, 8, 8, 9, 5, 8, 8, 5, 8, 9, 7, 8, 12, 6, 7, 10, 8, 9, 9, 7, 8, 11, 12, 8, 8, 10, 8, 7, 6, 10, 10, 9, 7, 10, 9, 7, 6, 9

Offset: 1

N. J. A. Sloane

a(n) is the maximal number of positive reduced quadratic nonresidues which appear consecutively for the n-th prime.

When prime(n) == 3 (mod 4), then a(n) = A002307(n). - T. D. Noe, Apr 03 2007

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
A. A. Bennett, Consecutive quadratic residues, Bull. Amer. Math. Soc., 32 (1926), 283-284.

Cf. A002307.

Cf. A129201

f[l_, a_] := Module[{A = Split[l], B}, B = Last[Sort[ Cases[A, x : {a ..} :> {Length[x], Position[A, x][[1, 1]]}]]]; {First[B], Length[Flatten[Take[A, Last[B] - 1]]] + 1}]; g[n_] := f[-JacobiSymbol[Range[Prime[n] - 1], Prime[n]], 1][[1]]; g[1] = 0; Table[g[n], {n, 1, 107}] (* Jean-François Alcover, Oct 17 2012, after the Mathematica code of Robert G. Wilson v in A002307 *)

More terms from David W. Wilson

A048280 Length of longest run of consecutive quadratic residues mod prime(n).

Original entry on oeis.org

2, 2, 3, 3, 3, 3, 5, 4, 5, 4, 4, 4, 5, 5, 5, 3, 5, 5, 6, 7, 9, 6, 7, 5, 9, 7, 7, 6, 5, 5, 7, 8, 6, 5, 4, 7, 6, 6, 6, 6, 6, 6, 7, 9, 7, 6, 7, 7, 7, 5, 6, 7, 13, 7, 6, 7, 8, 7, 10, 6, 9, 9, 7, 11, 9, 5, 8, 9, 8, 6, 6, 8, 9, 6, 8, 8, 8, 5, 7, 13, 8, 7, 7, 9, 10, 8, 8, 9, 8, 8, 11, 13, 8, 8, 10, 8, 9, 8, 10, 7, 9, 9, 10, 10, 7, 9

Offset: 1

David W. Wilson

nonn

0 and 1 are consecutive quadratic residues for any prime, so a(n) >= 2.
"Consecutive" allows wrap-around, so p-1 and 0 are consecutive. - Robert Israel, Jul 20 2014
A002307(n) is defined similarly, except that only positive reduced quadratic residues are counted. - Jonathan Sondow, Jul 20 2014
For longest runs of quadratic nonresidues, see A002308. - Jonathan Sondow, Jul 20 2014

			For n = 7, prime(7) = 17 has consecutive quadratic residues 15,16,0,1,2, and no longer sequence of consecutive quadratic residues, so a(7)=5.

Robert Israel, Table of n, a(n) for n = 1..10000
P. Pollack and E. Treviño, The primes that Euclid forgot, Amer. Math. Monthly, 121 (2014), 433-437.
Enrique Treviño, Corrigendum to “On the maximum number of consecutive integers on which a character is constant”, Mosc. J. Comb. Number Theory 7 (2017), no. 3, 1-2.

Cf. A002307, A048281.

A:= proc(n) local P, res, nonres, nnr;
     P:= ithprime(n);
     res:= {seq(i^2,i=0..floor((P-1)/2)};
     nonres:= {$1..P-1} minus res;
     nnr:= nops(nonres);
     max(seq(nonres[i+1]-nonres[i]-1,i=1..nnr-1),nonres[1]-nonres[-1]+P-1)
end proc;
A(1):= 2:
seq(A(n),n=1..100); # Robert Israel, Jul 20 2014

a(n) < 2*sqrt(prime(n)) for n >= 1 (see Pollack and Treviño for n > 1). - Jonathan Sondow, Jul 20 2014
a(n) >= A002307(n). - Jonathan Sondow, Jul 20 2014
a(n) < 7 prime(n)^(1/4)log(prime(n)) for all n > 1, or a(n) < 3.2 prime(n)^(1/4)log(prime(n)) for n >= 10^13. - Enrique Treviño, Apr 16 2020

Offset corrected to 1 and definition clarified by Jonathan Sondow Jul 20 2014

A125607 Lesser of the smallest pair of consecutive positive reduced quadratic residues modulo p = prime(n) > 5.

Original entry on oeis.org

1, 3, 3, 1, 4, 1, 4, 1, 3, 1, 9, 1, 6, 3, 3, 9, 1, 1, 1, 3, 1, 1, 4, 1, 3, 3, 1, 1, 3, 1, 4, 4, 1, 3, 9, 1, 9, 3, 3, 1, 1, 6, 1, 4, 1, 3, 3, 1, 1, 1, 3, 1, 1, 4, 1, 3, 1, 6, 9, 6, 1, 1, 6, 4, 1, 3, 3, 1, 1, 1, 3, 4, 1, 4, 3, 1, 1, 3, 3, 1, 1, 1, 3, 1, 1, 4, 1, 3, 1, 1, 3, 4, 1, 4, 1, 6, 3, 9, 6, 3, 1, 4, 1, 3, 1

Offset: 4

Nick Hobson, Nov 30 2006

For all n, a(n) exists and equals 1, 3, 4, 6 or 9. Proof: a(4)=1 by inspection. For n > 4 (p > 7), if 2 is a quadratic residue of p, then a(n)=1; otherwise if 5 is a quadratic residue of p, then a(n)=4 or 3; otherwise 2*5=10 is a quadratic residue of p and (9, 10) are consecutive residues. However, a(n)=8 or 7 is impossible as 8 cannot be a quadratic residue (since 2 is not), leaving 9 and 6 as the other possible values.

The constant 0.133141413191633911131141331131441391... = sum(a(n)/10^(n-3)) is conjectured to be irrational.

			The quadratic residues of 13=prime(6) are 1, 3, 4, 9, 10 and 12. The least consecutive pair of residues is (3, 4); hence a(6)=3.

Nick Hobson, Table of n, a(n) for n = 4..1003
D. H. Lehmer and Emma Lehmer, On Runs of Residues, Proc. American Mathematical Society, Vol. 13, No. 1 (Feb., 1962), pp. 102-106.

Cf. A000445, A002307, A097159, A097160, A097161.

vector(108, m, p=prime(m+3); if(p%8==1||p%8==7, 1, if(p%12==1||p%12==11, 3, if(p%10==1||p%10==9, 4, if((p%24==1||p%24==5||p%24==19||p%24==23) && (p%28==1||p%28==3||p%28==9||p%28==19||p%28==25||p%28==27), 6, 9)))))

cp's OEIS Frontend

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

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A097159 Smallest prime p such that there are n consecutive quadratic residues mod p.

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A002308 Consecutive quadratic nonresidues mod p.

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A048280 Length of longest run of consecutive quadratic residues mod prime(n).

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A125607 Lesser of the smallest pair of consecutive positive reduced quadratic residues modulo p = prime(n) > 5.

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A007070 a(n) = 4a(n-1) - 2a(n-2) with a(0) = 1, a(1) = 4.