cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 26 results. Next

A104414 Number of prime factors, with multiplicity, of the heptanacci numbers A066178.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 5, 6, 1, 2, 6, 3, 7, 6, 9, 8, 1, 4, 2, 5, 6, 5, 8, 9, 2, 3, 10, 6, 7, 7, 16, 10, 4, 2, 7, 5, 6, 9, 12, 10, 4, 3, 6, 4, 9, 8, 14, 12, 2, 3, 7, 6, 11, 8, 7, 10, 5, 5, 12, 6, 7, 9, 12, 11, 3, 4, 3, 6, 7, 5, 6, 11, 4, 2, 9, 4, 7, 9, 14, 8, 4, 3
Offset: 0

Views

Author

Jonathan Vos Post, Mar 06 2005

Keywords

Comments

Prime heptanacci numbers: a(2) = 2, a(8) = 127, a(16) = 31489, ... Semiprime heptanacci numbers: a(4) = 4 = 2^2, a(9) = 253 = 11 * 23, a(18) = 124946 = 2 * 62473, a(24) = 7805695 = 5 * 1561139.

Examples

			a(0)=a(1)=0 because the first two nonzero heptanacci numbers are both 1, which has zero prime divisors.
a(2)=1 because the 3rd nonzero heptanacci number is 2, a prime, with only one prime divisor.
a(3)=2 because the 4th nonzero pentanacci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).
a(4)=3 because the 5th nonzero heptanacci number is 8 = 2^3.
a(12)= 7 because A066178(12) = 2000 = 2^4 * 5^3 which has seven prime factors (four of the 2, three of them 5).
		

Crossrefs

Programs

  • Mathematica
    PrimeOmega[#]&/@LinearRecurrence[{1,1,1,1,1,1,1},{1,1,2,4,8,16,32},100] (* Harvey P. Dale, Oct 08 2015 *)

Formula

a(n) = A001222(A066178(n)). a(n) = bigomega(A066178(n)).

Extensions

More terms from Harvey P. Dale, Oct 08 2015

A105761 Prime Fibonacci 7-step numbers, A066178.

Original entry on oeis.org

2, 127, 31489
Offset: 1

Views

Author

T. D. Noe, Apr 22 2005, May 13 2008

Keywords

Comments

The next term is the 30673-digit probable prime 59456197869789581...206057955817 that is the 102490th term of the heptanacci sequence (using the definition in Noe-Post and Weisstein). This is too large to include here.

Programs

  • Mathematica
    a={1, 0, 0, 0, 0, 0, 0}; lst={}; Do[s=Plus@@a; a=RotateLeft[a]; a[[ -1]]=s; If[PrimeQ[s], AppendTo[lst, s]], {n, 100}]; lst

A079262 Octanacci numbers: a(0)=a(1)=...=a(6)=0, a(7)=1; for n >= 8, a(n) = Sum_{i=1..8} a(n-i).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, 1016, 2028, 4048, 8080, 16128, 32192, 64256, 128257, 256005, 510994, 1019960, 2035872, 4063664, 8111200, 16190208, 32316160, 64504063, 128752121, 256993248, 512966536, 1023897200, 2043730736
Offset: 0

Views

Author

Michael Joseph Halm, Feb 04 2003

Keywords

Comments

a(n+7) is the number of compositions of n into parts <= 8. - Joerg Arndt, Sep 24 2020

Examples

			a(16) = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255.
		

Crossrefs

Row 8 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Cf. A253706, A253705. Primes and indices of primes in this sequence.

Programs

  • Maple
    for j from 0 to 6 do a[j]:=0 od: a[7]:=1: for n from 8 to 45 do a[n]:=sum(a[n-i],i=1..8) od:seq(a[n],n=0..45); # Emeric Deutsch, Apr 16 2005
  • Mathematica
    LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    With[{nn=8},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)

Formula

G.f.: x^7/(1 - x - x^2 - x^3 - x^4 - x^5 - x^6 - x^7 - x^8). - Emeric Deutsch, Apr 16 2005
a(1)..a(9) = 1, 1, 2, 4, 8, 16, 32, 64, 128. a(10) and following are given by 63*2^(n-8)+(1/2+sqrt(5/4))^(n-6)/sqrt(5)-(1/2-sqrt(5/4))^(n-6)/sqrt(5). Offset 10. a(10)=255. - Al Hakanson (hawkuu(AT)gmail.com), Feb 14 2009
Another form of the g.f.: f(z) = (z^7 - z^8)/(1 - 2*z + z^9), then a(n) = Sum_{i=0..floor((n-7)/9)} (-1)^i*binomial(n-7-8*i,i)*2^(n-7-9*i) - Sum_{i=0..floor((n-8)/9)} (-1)^i*binomial(n-8-8*i,i)*2^(n-8-9*i) with Sum_{i=m..n} alpha(i) = 0 for m>n. - Richard Choulet, Feb 22 2010
Sum_{k=0..7*n} a(k+b)*A171890(n,k) = a(8*n+b), b>=0.
a(n) = 2*a(n-1) - a(n-9). - Vincenzo Librandi, Dec 20 2010

Extensions

Corrected by Joao B. Oliveira (oliveira(AT)inf.pucrs.br), Nov 25 2004

A092921 Array F(k, n) read by descending antidiagonals: k-generalized Fibonacci numbers in row k >= 1, starting (0, 1, 1, ...), for column n >= 0.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 2, 1, 1, 0, 1, 5, 4, 2, 1, 1, 0, 1, 8, 7, 4, 2, 1, 1, 0, 1, 13, 13, 8, 4, 2, 1, 1, 0, 1, 21, 24, 15, 8, 4, 2, 1, 1, 0, 1, 34, 44, 29, 16, 8, 4, 2, 1, 1, 0, 1, 55, 81, 56, 31, 16, 8, 4, 2, 1, 1, 0, 1, 89, 149, 108, 61, 32, 16, 8, 4, 2, 1, 1, 0
Offset: 0

Views

Author

Ralf Stephan, Apr 17 2004

Keywords

Comments

For all k >= 1, the k-generalized Fibonacci number F(k,n) satisfies the recurrence obtained by adding more terms to the recurrence of the Fibonacci numbers.
The number of tilings of an 1 X n rectangle with tiles of size 1 X 1, 1 X 2, ..., 1 X k is F(k,n).
T(k,n) is the number of 0-balanced ordered trees with n edges and height k (height is the number of edges from root to a leaf). - Emeric Deutsch, Jan 19 2007
Brlek et al. (2006) call this table "number of psp-polyominoes with flat bottom". - N. J. A. Sloane, Oct 30 2018

Examples

			From _Peter Luschny_, Apr 03 2021: (Start)
Array begins:
                n = 0  1  2  3  4  5   6   7   8    9   10
  -------------------------------------------------------------
  [k=1, mononacci ] 0, 1, 1, 1, 1, 1,  1,  1,  1,   1,   1, ...
  [k=2, Fibonacci ] 0, 1, 1, 2, 3, 5,  8, 13, 21,  34,  55, ...
  [k=3, tribonacci] 0, 1, 1, 2, 4, 7, 13, 24, 44,  81, 149, ...
  [k=4, tetranacci] 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, ...
  [k=5, pentanacci] 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, ...
  [k=6]             0, 1, 1, 2, 4, 8, 16, 32, 63, 125, 248, ...
  [k=7]             0, 1, 1, 2, 4, 8, 16, 32, 64, 127, 253, ...
  [k=8]             0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 255, ...
  [k=9]             0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, ...
Note that the first parameter in F(k, n) refers to rows, and the second parameter refers to columns. This is always the case. Only the usual naming convention for the indices is not adhered to because it is common to call the row sequences k-bonacci numbers. (End)
.
From _Peter Luschny_, Aug 12 2015: (Start)
As a triangle counting compositions of n with largest part k:
  [n\k]| [0][1] [2] [3] [4][5][6][7][8][9]
   [0] | [0]
   [1] | [0, 1]
   [2] | [0, 1,  1]
   [3] | [0, 1,  1,  1]
   [4] | [0, 1,  2,  1,  1]
   [5] | [0, 1,  3,  2,  1, 1]
   [6] | [0, 1,  5,  4,  2, 1, 1]
   [7] | [0, 1,  8,  7,  4, 2, 1, 1]
   [8] | [0, 1, 13, 13,  8, 4, 2, 1, 1]
   [9] | [0, 1, 21, 24, 15, 8, 4, 2, 1, 1]
For example for n=7 and k=3 we have the 7 compositions [3, 3, 1], [3, 2, 2], [3, 2, 1, 1], [3, 1, 3], [3, 1, 2, 1], [3, 1, 1, 2], [3, 1, 1, 1, 1]. (End)
		

Crossrefs

Columns converge to A166444: each column n converges to A166444(n) = 2^(n-2).
Rows 1-8 are (shifted) A057427, A000045, A000073, A000078, A001591, A001592, A066178, A079262.
Essentially a reflected version of A048887.
See A048004 and A126198 for closely related arrays.
Cf. A066099.

Programs

  • Maple
    F:= proc(k, n) option remember; `if`(n<2, n,
          add(F(k, n-j), j=1..min(k,n)))
        end:
    seq(seq(F(k, d+1-k), k=1..d+1), d=0..12);  # Alois P. Heinz, Nov 02 2016
    # Based on the above function:
    Arow := (k, len) -> seq(F(k, j), j = 0..len):
    seq(lprint(Arow(k, 14)), k = 1..10); # Peter Luschny, Apr 03 2021
  • Mathematica
    F[k_, n_] := F[k, n] = If[n<2, n, Sum[F[k, n-j], {j, 1, Min[k, n]}]];
    Table[F[k, d+1-k], {d, 0, 12}, {k, 1, d+1}] // Flatten (* Jean-François Alcover, Jan 11 2017, translated from Maple *)
  • PARI
    F(k,n)=if(n<2,if(n<1,0,1),sum(i=1,k,F(k,n-i)))
    
  • PARI
    T(m,n)=!!n*(matrix(m,m,i,j,j==i+1||i==m)^(n+m-2))[1,m] \\ M. F. Hasler, Apr 20 2018
    
  • PARI
    F(k,n) = if(n==0,0, polcoeff(lift(Mod('x, Pol(vector(k+1,i, if(i==1,1,-1))))^(n+k-2)), k-1)); \\ Kevin Ryde, Jun 05 2020
    
  • Sage
    # As a triangle of compositions of n with largest part k.
    C = lambda n,k: Compositions(n, max_part=k, inner=[k]).cardinality()
    for n in (0..9): [C(n,k) for k in (0..n)] # Peter Luschny, Aug 12 2015

Formula

F(k,n) = F(k,n-1) + F(k,n-2) + ... + F(k,n-k); F(k,1) = 1 and F(k,n) = 0 for n <= 0.
G.f.: x/(1-Sum_{i=1..k} x^i).
F(k,n) = 2^(n-2) for 1 < n <= k+1. - M. F. Hasler, Apr 20 2018
F(k,n) = Sum_{j=0..floor(n/(k+1))} (-1)^j*((n - j*k) + j + delta(n,0))/(2*(n - j*k) + delta(n,0))*binomial(n - j*k, j)*2^(n-j*(k+1)), where delta denotes the Kronecker delta (see Corollary 3.2 in Parks and Wills). - Stefano Spezia, Aug 06 2022

A104144 a(n) = Sum_{k=1..9} a(n-k); a(8) = 1, a(n) = 0 for n < 8.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512, 64960, 129792, 259328, 518145, 1035269, 2068498, 4132920, 8257696, 16499120, 32965728, 65866496, 131603200, 262947072, 525375999, 1049716729, 2097364960
Offset: 0

Views

Author

Jean Lefort (jlefort.apmep(AT)wanadoo.fr), Mar 07 2005

Keywords

Comments

Sometimes called the Fibonacci 9-step numbers.
For n >= 8, this gives the number of integers written without 0 in base ten, the sum of digits of which is equal to n-7. E.g., a(11) = 8 because we have the 8 numbers: 4, 13, 22, 31, 112, 121, 211, 1111.
The offset for this sequence is fairly arbitrary. - N. J. A. Sloane, Feb 27 2009

Crossrefs

Cf. A000045, A000073, A000078, A001591, A001592, A066178, A079262 (Fibonacci n-step numbers).
Cf. A255529 (Indices of primes in this sequence).

Programs

  • Maple
    for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-8-9*i,i)*2^(n-8-10*i),i=0..floor((n-8)/10))-sum((-1)^i*binomial(n-9-9*i,i)*2^(n-9-10*i),i=0..floor((n-9)/10)):od:seq(k(n),n=0..50);a:=taylor((z^8-z^9)/(1-2*z+z^(10)),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010
  • Mathematica
    a={1, 0, 0, 0, 0, 0, 0, 0, 0}; Table[s=Plus@@a; a=RotateLeft[a]; a[[ -1]]=s, {n, 50}]
    LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    With[{nn=9},LinearRecurrence[Table[1,{nn}],Join[Table[0,{nn-1}],{1}],50]] (* Harvey P. Dale, Aug 17 2013 *)
  • PARI
    a(n)=([0,1,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0; 0,0,0,0,1,0,0,0,0; 0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,1; 1,1,1,1,1,1,1,1,1]^n*[0;0;0;0;0;0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 16 2015
    
  • PARI
    A104144(n,m=9)=(matrix(m,m,i,j,j==i+1||i==m)^n)[1,m] \\ M. F. Hasler, Apr 22 2018

Formula

a(n) = Sum_{k=1..9} a(n-k) for n > 8, a(8) = 1, a(n) = 0 for n=0..7.
G.f.: x^8/(1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9). - N. J. A. Sloane, Dec 04 2011
Another form of the g.f. f: f(z) = (z^8-z^9)/(1-2*z+z^(10)), then a(n) = Sum_((-1)^i*binomial(n-8-9*i,i)*2^(n-8-10*i), i=0..floor((n-8)/10))-Sum_((-1)^i*binomial(n-9-9*i,i)*2^(n-9-10*i), i=0..floor((n-9)/10)) with Sum_(alpha(i), i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010
From N. J. A. Sloane, Dec 04 2011: (Start)
Let b be the smallest root (in magnitude) of g(x) := 1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9, b = 0.50049311828655225605926845999420216157202861343888...
Let c = -b^8/g'(b) = 0.00099310812055463178382193226558248643030626601288701...
Then a(n) is the nearest integer to c/b^n. (End)

Extensions

Edited by N. J. A. Sloane, Aug 15 2006 and Nov 11 2006
Incorrect formula deleted by N. J. A. Sloane, Dec 04 2011
Name edited by M. F. Hasler, Apr 22 2018

A104621 Heptanacci-Lucas numbers.

Original entry on oeis.org

7, 1, 3, 7, 15, 31, 63, 127, 247, 493, 983, 1959, 3903, 7775, 15487, 30847, 61447, 122401, 243819, 485679, 967455, 1927135, 3838783, 7646719, 15231991, 30341581, 60439343, 120393007, 239818559, 477709983, 951581183, 1895515647, 3775799303, 7521257025
Offset: 0

Views

Author

Jonathan Vos Post, Mar 17 2005

Keywords

Comments

This 7th-order linear recurrence is a generalization of the Lucas sequence A000032. Mario Catalani would refer to this is a generalized heptanacci sequence, had he not stopped his series of sequences after A001644 "generalized tribonacci", A073817 "generalized tetranacci", A074048 "generalized pentanacci", A074584 "generalized hexanacci." T. D. Noe and I have noted that each of these has many more primes than the corresponding tribonacci A000073 (see A104576), tetranacci A000288 (see A104577), pentanacci, hexanacci and heptanacci (see A104414). For primes in Heptanacci-Lucas numbers, see A104622. For semiprimes in Heptanacci-Lucas numbers, see A104623.

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (-7+6*x+ 5*x^2+4*x^3+3*x^4+2*x^5+x^6)/(-1+x +x^2+x^3+x^4+x^5+x^6+x^7) )); // G. C. Greubel, Apr 22 2019
    
  • Maple
    A104621 := proc(n)
        option remember;
        if n <=6 then
            op(n+1,[7, 1, 3, 7, 15, 31, 63])
        else
            add(procname(n-i),i=1..7) ;
        end if;
    end proc: # R. J. Mathar, Mar 26 2015
  • Mathematica
    a[0]=7; a[1]=1; a[2]=3; a[3]=7; a[4]=15; a[5]=31; a[6]=63; a[n_]:= a[n]= a[n-1]+a[n-2]+a[n-3]+a[n-4]+a[n-5]+a[n-6]+a[n-7]; Table[a[n], {n,0,40}] (* Robert G. Wilson v, Mar 17 2005 *)
    LinearRecurrence[{1, 1, 1, 1, 1, 1, 1}, {7, 1, 3, 7, 15, 31, 63}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
  • PARI
    my(x='x+O('x^40)); Vec((-7+6*x+5*x^2+4*x^3+3*x^4+2*x^5+x^6)/(-1+x +x^2+x^3+x^4+x^5+x^6+x^7)) \\ G. C. Greubel, Dec 18 2017
    
  • PARI
    polsym(polrecip(1-x-x^2-x^3-x^4-x^5-x^6-x^7), 40) \\ G. C. Greubel, Apr 22 2019
    
  • Sage
    ((-7+6*x+5*x^2+4*x^3+3*x^4+2*x^5+x^6)/(-1+x +x^2+x^3+x^4+x^5+x^6 +x^7)).series(x, 41).coefficients(x, sparse=False) # G. C. Greubel, Apr 22 2019

Formula

a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) + a(n-6) + a(n-7); a(0) = 7, a(1) = 1, a(2) = 3, a(3) = 7, a(4) = 15, a(5) = 31, a(6) = 63.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: (7 - 6*x - 5*x^2 - 4*x^3 - 3*x^4 - 2*x^5 - x^6)/(1 - x - x^2 - x^3 - x^4 - x^5 - x^6 - x^7).
a(n) = 7*A066178(n) - 6*A066178(n-1) - 5*A066178(n-2) - ... - 2*A066178(n-5) - A066178(n-6) if n >= 6. (End)

A122189 Heptanacci numbers: each term is the sum of the preceding 7 terms, with a(0),...,a(6) = 0,0,0,0,0,0,1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, 1004, 2000, 3984, 7936, 15808, 31489, 62725, 124946, 248888, 495776, 987568, 1967200, 3918592, 7805695, 15548665, 30972384, 61695880, 122895984, 244804400, 487641600, 971364608, 1934923521
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Oct 18 2006

Keywords

Comments

See A066178 (essentially the same sequence) for more about the heptanacci numbers and other generalizations of the Fibonacci numbers (A000045).

Crossrefs

Cf. A000045 (k=2, Fibonacci numbers), A000073 (k=3, tribonacci) A000078 (k=4, tetranacci) A001591 (k=5, pentanacci) A001592 (k=6, hexanacci), A122189 (k=7, heptanacci).

Programs

  • Maple
    for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i),i=0..floor((n-6)/8))-sum((-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i),i=0..floor((n-7)/8)):od:seq(k(n),n=0..50); a:=taylor((z^6-z^7)/(1-2*z+z^8),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010
  • Mathematica
    LinearRecurrence[{1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    a={0,0,0,0,0,0,1} For[n=7, n≤100, n++, sum=Plus@@a; Print[sum]; a=RotateLeft[a]; a[[7]]=sum] (* Robert Price, Dec 04 2014 *)
  • PARI
    a(n)=([0,1,0,0,0,0,0; 0,0,1,0,0,0,0; 0,0,0,1,0,0,0; 0,0,0,0,1,0,0; 0,0,0,0,0,1,0; 0,0,0,0,0,0,1; 1,1,1,1,1,1,1]^n*[0;0;0;0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Formula

G.f.: x^6/(1-x-x^2-x^3-x^4-x^5-x^6-x^7). - R. J. Mathar, Feb 13 2009
G.f.: Sum_{n >= 0} x^(n+5) * [ Product_{k = 1..n} (k + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + x^6)/(1 + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + k*x^6) ]. - Peter Bala, Jan 04 2015
Another form of the g.f.: f(z) = (z^6-z^7)/(1-2*z+z^8), then a(n) = Sum_{i=0..floor((n-6)/8)} (-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i) - Sum_{i=0..floor((n-7)/8)} (-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i) with Sum_{i=m..n} alpha(i) = 0 for m>n. - Richard Choulet, Feb 22 2010
Sum_{k=0..6*n} a(k+b)*A063265(n,k) = a(7*n+b), b>=0.
a(n) = 2*a(n-1) - a(n-8). - Joerg Arndt, Sep 24 2020

Extensions

Edited by N. J. A. Sloane, Nov 20 2007
Wrong Binet-type formula removed by R. J. Mathar, Feb 13 2009

A107245 Sum of squares of heptanacci numbers (Fibonacci 7-step numbers).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21591, 85600, 339616, 1347632, 5347632, 21219888, 84199984, 334092848, 1325649969, 5260075594, 20871578510, 82816815054, 328610657230, 1303901211854, 5173777051854, 20529140314318
Offset: 0

Views

Author

Jonathan Vos Post, May 20 2005

Keywords

Comments

Primes include: a(7) = 2. Semiprimes include a(8) = 6 = 2 * 3, a(9) = 22 = 2 * 11, a(10) = 86 = 2 * 43, a(12) = 1366 = 2 * 683, a(13) = 5462 = 2 * 2731.

Examples

			a(0) = 0 = 0^2
a(1) = 0 = 0^2 + 0^2
a(2) = 0 = 0^2 + 0^2 + 0^2
a(3) = 0 = 0^2 + 0^2 + 0^2 + 0^2
a(4) = 0 = 0^2 + 0^2 + 0^2 + 0^2 + 0^2
a(5) = 0 = 0^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2
a(6) = 1 = 0^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2 + 1^2
a(7) = 2 = 0^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2 + 1^2 + 1^2
a(8) = 6 = 0^2 + 0^2 + 0^2+ 0^2 + 0^2 + 0^2 + 1^2 + 1^2 + 2^2
a(9) = 22 = 0^2 + 0^2 +0^2 + 0^2 + 0^2 + 1^2 + 1^2 + 2^2 + 4^2 = 2*11
a(10) = 86 = 8^2 + 22
a(11) = 342 = 16^2 + 86
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, 2, 4, 7, 15, 30, 60, -131, -9, -8, 28, -11, -25, -32, 68, 5, 5, -10, 0, 5, 9, -14, 0, -1, 1, 0, 0, -1, 1},{0, 0, 0, 0, 0, 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21591, 85600, 339616, 1347632, 5347632, 21219888, 84199984, 334092848, 1325649969, 5260075594, 20871578510, 82816815054, 328610657230, 1303901211854, 5173777051854},30] (* Ray Chandler, Aug 02 2015 *)
    Accumulate[LinearRecurrence[{1,1,1,1,1,1,1},{0,0,0,0,0,0,1},30]^2] (* Ray Chandler, Aug 02 2015 *)

Formula

a(n) = F_7(0)^2 + F_7(1)^2 + ... F_7(n)^2, note that F_7(n) = A066178(n) with corrected offset (from leading zeros). a(0) = 0, a(n+1) = a(n) + F_7(n)^2.

Extensions

a(14) inserted by R. J. Mathar, Aug 11 2009

A118428 Decimal expansion of heptanacci constant.

Original entry on oeis.org

1, 9, 9, 1, 9, 6, 4, 1, 9, 6, 6, 0, 5, 0, 3, 5, 0, 2, 1, 0, 9, 7, 7, 4, 1, 7, 5, 4, 5, 8, 4, 3, 7, 4, 9, 6, 3, 4, 7, 9, 3, 1, 8, 9, 6, 0, 0, 5, 3, 1, 5, 7, 9, 9, 5, 2, 4, 4, 7, 8, 2, 1, 5, 3, 4, 0, 0, 9, 5, 1, 9, 8, 0, 3, 0, 9, 6, 2, 2, 1, 8, 3, 5, 6, 3, 1, 4, 1, 5, 7, 7, 0, 2, 2, 7, 1, 9, 0, 1, 7, 0, 9, 9, 1, 6
Offset: 1

Views

Author

Eric W. Weisstein, Apr 27 2006

Keywords

Comments

Other roots of the equation x^7 - x^6 - ... - x - 1 see in A239566. For n>=7, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-7) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 07 2022

Examples

			1.9919641966050350210...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.

Crossrefs

k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), A086088 (tetranacci), A103814 (pentanacci), A118427 (hexanacci), this sequence (heptanacci).

Programs

  • Mathematica
    RealDigits[x/.FindRoot[x^7+Total[-x^Range[0,6]]==0,{x,2}, WorkingPrecision-> 110]][[1]] (* Harvey P. Dale, Dec 13 2011 *)
  • PARI
    polrootsreal(x^7 - x^6 - x^5 - x^4 - x^3 - x^2 - x - 1)[1] \\ Charles R Greathouse IV, Feb 11 2025

A172317 8th column of A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 255, 508, 1012, 2016, 4016, 8000, 15936, 31744, 63233, 125958, 250904, 499792, 995568, 1983136, 3950336, 7868928, 15674623, 31223288, 62195672, 123891552, 246787536, 491591936, 979233536
Offset: 0

Views

Author

Richard Choulet, Jan 31 2010

Keywords

Examples

			a(4) = binomial(4,4)*2^4 = 16.
a(9) = binomial(9,9)*2^9 - binomial(2,1)*2^1 = 512 - 4 = 508.
		

Crossrefs

Partial sums of A066178.

Programs

  • Maple
    k:=7:taylor(1/(1-2*z+z^(k+1)),z=0,30); for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od;

Formula

The generating function is f such that: f(z)=1/(1-2*z+z^8). Recurrence relation: a(n+8)=2*a(n+7)-a(n). General term: a(n) = Sum_{j=0..floor(n/(k+1))} ((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)) with k=7.
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