Ruskin Harding - cp's OEIS frontend

A220469 Fibonacci 14-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-14).

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32765, 65528, 131052, 262096, 524176, 1048320, 2096576, 4193024, 8385792, 16771072, 33541120, 67080192, 134156288, 268304384, 536592385, 1073152005, 2146238482, 4292345912, 8584429728

Offset: 1

Ruskin Harding, Feb 20 2013

nonn

Also called tetradecanacci numbers. In previous similar sequences, a(1),...,a(n-1) have been set equal to zero and a(n)=1. For example, A168084 (Fibonacci 13-step numbers) has 12 0's as the first 12 terms and a(13)=1.

M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4
Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1).

Cf. A000045 (Fibonacci), A000073 (tribonacci), A000078 (tetranacci), A001591 (pentanacci).

FibonacciSequence[n_, kMax_] := Module[{a, s}, a = Join[{1}, Table[0, {n - 1}]]; lst = {}; Table[s = Plus @@ a; a = RotateLeft[a]; a[[n]] = s, {k, 1, kMax}]]; FibonacciSequence[14, 50] (* T. D. Noe, Feb 20 2013 *)
Drop[LinearRecurrence[PadRight[{},14,1],Join[PadRight[{},13,0],{1}],50],13] (* Harvey P. Dale, Feb 25 2013 *)
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},{1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096},35] (* Ray Chandler, Aug 03 2015 *)

A220493 Fibonacci 15-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-15).

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65533, 131064, 262124, 524240, 1048464, 2096896, 4193728, 8387328, 16774400, 33548288, 67095552, 134189056, 268374016, 536739840, 1073463296, 2146893825, 4293722117, 8587313170

Offset: 1

Ruskin Harding, Feb 20 2013

Also called Pentadecanacci numbers. In previous similar sequences, a(1), ..., a(n-1) have been set equal to zero and a(n)=1. For example, A168084 (Fibonacci 13-step numbers) has 12 0's as the first 12 terms and a(13)=1.

Robert Israel, Table of n, a(n) for n = 1..3320
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1).

Cf. A000045 (Fibonacci), A000073 (tribonacci), A000078 (tetranacci), A001591 (pentanacci).

f:= gfun:-rectoproc({a(n) = add(a(n-i),i=1..15), seq(a(n)=0,n=-14..0),a(1)=1},a(n),remember):
map(f, [$1..100]); # Robert Israel, Feb 19 2019

FibonacciSequence[n_, kMax_] := Module[{a, s}, a = Join[{1}, Table[0, {n - 1}]]; lst = {}; Table[s = Plus @@ a; a = RotateLeft[a]; a[[n]] = s, {k, 1, kMax}]]; FibonacciSequence[15, 50] (* T. D. Noe, Feb 20 2013 *)

G.f.: x/(1-Sum_{k=1..15} x^k). - Robert Israel, Feb 19 2019

A192543 Let r be the largest real zero of x^n - x^(n-1) - x^(n-2) - ... - 1 = 0. Then a(n) is the value of k which satisfies the equation 0.5/10^k < 2 - r < 5/10^k.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22

Offset: 1

Ruskin Harding, Dec 31 2012

Same as A034887 except for the offset and a(1). - T. D. Noe, Feb 11 2013

			For n = 5, the root is approximately r = 1.96594823. The value of k that satisfies 0.5/10^k < 2-r < 5/10^k is 2 as 0.005 < 0.03405177 < 0.05. So a(5) = 2.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

a(n)=if(n>1, -log(4-2*solve(x=1.5,2,x^n-(1-x^n)/(1-x)))\log(10)+1, 0) \\ Charles R Greathouse IV, Jan 15 2013

A211191 List of odd values of k for which k^2+4 has a factor that is a square number larger than 1.

Original entry on oeis.org

11, 29, 39, 61, 89, 111, 139, 161, 189, 199, 211, 213, 239, 261, 289, 309, 311, 339, 361, 365, 367, 389, 393, 411, 439, 461, 489, 511, 521, 539, 561, 589, 611, 639, 647, 661, 689, 705, 711, 739, 759, 761, 789, 791, 811, 839, 861, 889, 911, 923, 925, 939, 943, 961, 985, 989

Offset: 1

Ruskin Harding, Feb 03 2013

nonn

Or, (odd integers n such that) n^2 + 4 is not squarefree. - Zak Seidov, Feb 03 2013

			The first odd value of k for which k^2+4 has a square factor is 11: 11^2+4 = 125 = 5^2*5.

Ruskin Harding, Table of n, a(n) for n = 1..5258

Cf. A000290, A005117, A005408, A087475.

[k: k in [1..1000 by 2] | not IsSquarefree(k^2+4)]; // Bruno Berselli, Feb 06 2013

Select[Range[11, 1000, 2], ! SquareFreeQ[#^2 + 4] &] (* Zak Seidov, Feb 03 2013 *)

is_term(n) = !issquarefree(n^2+4);
forstep (n=1,10^3,2, if (is_term(n), print1(n,", ")));
/* Joerg Arndt, Feb 05 2013 */

b=1
x=1
for i in range(1, 100000, 2):
    for j in range(2, i):
       if ((i**2)+4)%(j**2)==0:
          a=i
          if a!=b:
             b=a
             print(x, i)
             x=x+1

A187765 The (n-1)th decimal place of the fractional part of the square root of n.

Original entry on oeis.org

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

Offset: 1

Ruskin Harding, Jan 04 2013

If n=1 then the term is the first digit before the decimal point. If the square root of n is a whole number then the term is 0.

			If n=2, sqrt(2)=1.41421356 approx., the 1st(2-1) decimal place of which is 4 so the 2nd term is 4. If n=3, sqrt(3)=1.73205081 approx., the 2nd(3-1) decimal place of which is 3 so the 3rd term is 3.

Cf. A003076 (n-th digit after decimal point of square root of n).

Join[{0}, Table[RealDigits[Sqrt[n] - Floor[Sqrt[n]], 10, n, -1][[1, -2]], {n, 2, 87}]]

for n in range(1,16):
    x=str(n**0.5)
    for i in range(n):
        x=x+"0"
    if n==1:
        r=str(x[-1])
    else:
        r=r+","+str(x[n])
    if n==15:
        print(r)

A191244 a(1)=1; for n>1, a(n) = n*(10^n-1)/9 written in base n.

Original entry on oeis.org

1, 10110, 110100, 1011130, 3234210, 22142230, 123052510, 523053070, 2520607100, 11111111110, 479110292610, 19641010776330, 80101410369310, 210510130141118110, 1212103125891481260, 315281211715711270, 12138214138151315710, 56015814091312161590, 1780471401817161317410, 6151213981715112481715110

Offset: 1

Ruskin Harding, Dec 30 2012

For n = 2 through 9, this is the decimal number with n digits all equal to n, then written in base n. For n>9 the reader has to separate the "digits" himself (so this is a fairly unsatisfactory sequence).

A000461(n) written in base n.

			4444 in base 4 is 1011130.

f:=n->n*(10^n-1)/9;
g:=n->convert(f(n),base,n);
for n from 2 to 20 do
t1:=g(n);
t2:=nops(t1);
lprint( [seq(t1[t2+1-i],i=1..t2)]);

Join[{1}, Table[s = FromDigits[Table[n, {n}]]; FromDigits[IntegerDigits[s, n]], {n, 2, 9}]] (* T. D. Noe, Dec 30 2012 *)

Edited with Maple program by N. J. A. Sloane, Jan 05 2013

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User: Ruskin Harding

A220469 Fibonacci 14-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-14).

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A220493 Fibonacci 15-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-15).

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A192543 Let r be the largest real zero of x^n - x^(n-1) - x^(n-2) - ... - 1 = 0. Then a(n) is the value of k which satisfies the equation 0.5/10^k < 2 - r < 5/10^k.

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A211191 List of odd values of k for which k^2+4 has a factor that is a square number larger than 1.

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A187765 The (n-1)th decimal place of the fractional part of the square root of n.

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A191244 a(1)=1; for n>1, a(n) = n*(10^n-1)/9 written in base n.

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