A006449 -id:A006449 - cp's OEIS frontend

A006450 Prime-indexed primes: primes with prime subscripts.

3, 5, 11, 17, 31, 41, 59, 67, 83, 109, 127, 157, 179, 191, 211, 241, 277, 283, 331, 353, 367, 401, 431, 461, 509, 547, 563, 587, 599, 617, 709, 739, 773, 797, 859, 877, 919, 967, 991, 1031, 1063, 1087, 1153, 1171, 1201, 1217, 1297, 1409, 1433, 1447, 1471

Offset: 1

Jeffrey Shallit, Nov 25 1975

A000040 = A006450 U A007821. - Juri-Stepan Gerasimov, Sep 24 2009
Subsequence of A175247 (primes (A000040) with noncomposite (A008578) subscripts), a(n) = A175247(n+1). - Jaroslav Krizek, Mar 13 2010
Primes p such that p and pi(p) are both primes. - Juri-Stepan Gerasimov, Jul 14 2011
Sum_{n>=1} 1/a(n) converges. In fact, Sum_{n>N} 1/a(n) < 1/log(N), by the integral test. - Jonathan Sondow, Jul 11 2012
The number of such primes not exceeding x > 0 is pi(pi(x)). I conjecture that the sequence a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing. This is an analog of the Firoozbakht conjecture on primes. - Zhi-Wei Sun, Aug 17 2015
Limit_{n->infinity} a(n)/(n*(log(n))^2) = 1. Proof: By Cipolla's asymptotic formula, prime(n) ~ L(n) + R(n), where L(n)/n = log(n) + log(log(n)) - 1 and R(n)/n decreases logarithmically to 0. Hence, for large n, a(n) = prime(prime(n)) ~ L(L(n)+R(n)) + R(L(n)+R(n)) = n*(log(n))^2 + r(n), where r(n) grows as O(n*log(n)*log(log(n))). The rest of the proof is trivial. The convergence is very slow: for k = 1,2,3,4,5,6, sqrt(a(10^k)/10^k)/log(10^k) evaluates to 2.055, 1.844, 1.695, 1.611, 1.545, and 1.493, respectively. - Stanislav Sykora, Dec 09 2015

			a(5) = 31 because a(5) = p(p(5)) = p(11) = 31.

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

J. S. Kimberley, Table of n, a(n) for n = 1..100000
R. G. Batchko, A prime fractal and global quasi-self-similar structure in the distribution of prime-indexed primes, arXiv preprint arXiv:1405.2900 [math.GM], 2014.
Jonathan Bayless, Dominic Klyve, and Tomás Oliveira e Silva, New Bounds and Computations on Prime-Indexed Primes, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 13, Paper A43, 2013.
K. A. Broughan and A. R. Barnett, On the subsequence of primes having prime subscripts, JIS 12 (2009) 09.2.3.
Paul Cooijmans, Numbers.
Paul Cooijmans, Short Test For Genius.
R. E. Dressler and S. T. Parker, Primes with a prime subscript, J. ACM 22 (1975) 380-381.
N. Fernandez, An order of primeness, F(p)
N. Fernandez, An order of primeness [cached copy, included with permission of the author]
N. Fernandez, More terms of this and other sequences related to A049076.
A. B. Frizell, The permutations of the natural numbers can not be well ordered, Bull. Amer. Math. Soc. 22 (1915), no. 2, 71-73.
Ernest G. Hibbs, Component Interactions of the Prime Numbers, Ph. D. Thesis, Capitol Technology Univ. (2022), see p. 33.
Michael P. May, Properties of Higher-Order Prime Number Sequences, Missouri J. Math. Sci. (2020) Vol. 32, No. 2, 158-170; and arXiv version, arXiv:2108.04662 [math.NT], 2021.
Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Eric Weisstein's World of Mathematics, Prime formulas, see Cipolla formula.

Primes for which A049076 > 1.
Cf. A000040, A007821, A038580, A049090, A049202, A049203, A057847, A057849, A057850, A057851, A058332, A093047.
Cf. A185723 and A214296 for numbers and primes that are sums of distinct a(n); cf. A213356 and A185724 for those that are not.
Let A = primes A000040, B = nonprimes A018252. The 2-level compounds are AA = A006450, AB = A007821, BA = A078782, BB = A102615. The 3-level compounds AAA, AAB, ..., BBB are A038580, A049078, A270792, A102617, A270794, A270795, A270796, A102616.

a006450 = a000040 . a000040
a006450_list = map a000040 a000040_list
-- Reinhard Zumkeller, Jan 12 2013

[ NthPrime(NthPrime(n)): n in [1..51] ]; // Jason Kimberley, Apr 02 2010

seq(ithprime(ithprime(i)),i=1..50); # Uli Baum (Uli_Baum(AT)gmx.de), Sep 05 2007
# For Maple code for the prime/nonprime compound sequences (listed in cross-references) see A003622. - N. J. A. Sloane, Mar 30 2016

Mathematica
```
Table[ Prime[ Prime[ n ] ], {n, 100} ]
```

i=0;forprime(p=2,1e4,if(isprime(i++),print1(p", "))) \\ Charles R Greathouse IV, Jun 10 2011

a=vector(10^3,n,prime(prime(n))) \\ Stanislav Sykora, Dec 09 2015

from sympy import prime
def a(n): return prime(prime(n))
print([a(n) for n in range(1, 52)]) # Michael S. Branicky, Aug 11 2021

# much faster version for initial segment of sequence
from sympy import nextprime, isprime
def aupton(terms):
    alst, p, pi = [], 2, 1
    while len(alst) < terms:
        if isprime(pi): alst.append(p)
        p, pi = nextprime(p), pi+1
    return alst
print(aupton(10000)) # Michael S. Branicky, Aug 11 2021

a(n) = prime(prime(n)) = A000040(A000040(n)). - Juri-Stepan Gerasimov, Sep 24 2009
a(n) > n*(log(n))^2, as prime(n) > n*log(n) by Rosser's theorem. - Jonathan Sondow, Jul 11 2012
a(n)/log(a(n)) ~ prime(n). - Thomas Ordowski, Mar 30 2015
Sum_{n>=1} 1/a(n) is in the interval (1.04299, 1.04365) (Bayless et al., 2013). - Amiram Eldar, Oct 15 2020

A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

Original entry on oeis.org

0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152

Offset: 0

N. J. A. Sloane and Jeffrey Shallit

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Problem J/A2, Jul/August 1973 pp. 73-74; solution Jan 1974 (see link).
Jeffrey Shallit, personal communication.
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
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000124, A006452, A006454, A098586, A098790.

Cf. numbers m such that k*A000217(m)+1 is a square: this sequence for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

a006451 n = a006451_list !! n
a006451_list = 0 : 2 : 5 : 15 : map (+ 2)
   (zipWith (-) (map (* 6) (drop 2 a006451_list)) a006451_list)
-- Reinhard Zumkeller, Jan 10 2012

N:= 100: # to get a(0) to a(N)
A[0]:= 0: A[1]:= 2: A[2]:= 5: A[3]:= 15:
for n from 4 to N do A[n]:= 6*A[n-2] - A[n-4] + 2 od:
seq(A[n],n=0..N); # Robert Israel, Aug 26 2014

LinearRecurrence[{1,6,-6,-1,1},{0,2,5,15,32},30] (* Harvey P. Dale, Jul 17 2013 *)
Select[Range[10^6], IntegerQ@ Sqrt[# (# + 1)/2 + 1] &] (* Michael De Vlieger, Apr 25 2017 *)

for(n=1,10000,t=n*(n+1)/2+1;if(issquare(t), print1(n,", "))) \\ Joerg Arndt, Oct 10 2009

G.f.: x*(-2-3*x+2*x^2+x^3)/(x-1)/(x^2+2*x-1)/(x^2-2*x-1). Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a(n) = 6*a(n-2) - a(n-4) + 2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15. - Zak Seidov, Apr 15 2008
a(n) = 3*a(n-2) + 4*sqrt((a(n-2)^2 + a(n-2))/2 + 1) + 1 with a(0) = 0, a(1) = 2. - Raphie Frank, Feb 02 2013
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5); a(0)=0, a(1)=2, a(2)=5, a(3)=15, a(4)=32. - Harvey P. Dale, Jul 17 2013
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n+1) = A098790(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(2*n) = A098586(2*n-1), for n>0. - Hermann Stamm-Wilbrandt, Aug 27 2014
a(n) = 8*sqrt(T(a(n-2)) + 1) + a(n-4) where T(a(n)) = A000217(a(n)), and a(-1) = -1, a(0)=0, a(1)=2, a(2)=5. - Vladimir Pletser, Apr 29 2017

More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001

Edited by N. J. A. Sloane, Oct 24 2009, following discussions by several correspondents in the Sequence Fans Mailing List, Oct 10 2009

A006452 a(n) = 6*a(n-2) - a(n-4).

Original entry on oeis.org

1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552, 12671, 26531, 73852, 154634, 430441, 901273, 2508794, 5253004, 14622323, 30616751, 85225144, 178447502, 496728541, 1040068261, 2895146102, 6061962064, 16874148071, 35331704123

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Solution to a Diophantine equation.
Integers k such that k^2-1 is a triangular number. - Benoit Cloitre, Apr 05 2002
For all elements "x" of the sequence, 8*x^2 - 7 is a square. - Gregory V. Richardson, Oct 07 2002
a(n) mod 10 is a sequence of period 12: repeat (1, 1, 2, 4, 1, 3, 4, 4, 3, 1, 4, 2). - Paul Curtz, Dec 07 2012
a(n)^2 - 1 = A006454(n - 1) is a Sophie Germain triangular number of the second kind as defined in A217278. - Raphie Frank, Feb 08 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6xy + y^2 + 7 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 34xy + y^2 + 252 = 0. - Colin Barker, Mar 04 2014
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n+1), for n >= 0, gives one half of all positive y solutions of the Pell equation x^2 - 2*y^2 = -7. The corresponding x-solutions are x(n) = A077446(n+1).
See a comment on A077446 for the first and second class solutions separately, and the connection to the Pell equation X^2 - 2*Y^2 = 14. (End)
For n > 0, a(n) is the n-th almost balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: 11^2 - 2*(2*4)^2 = -7 (see the Pell comment above);
(4*4)^2 - 2*11^2 = +14. - _Wolfdieter Lang_, Feb 26 2015

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Jul/Aug 1973 pp. 73-74.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001333, A006451, A006452, A006454, A038723, A077446, A216134, A217278.

I:=[1,1,2,4]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013

A006452:=-(z-1)*(z**2+3*z+1)/(z**2+2*z-1)/(z**2-2*z-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

s=0;lst={1}; Do[s+=n;If[Sqrt[s+1]==Floor[Sqrt[s+1]],AppendTo[lst, Sqrt[s+1]]], {n,0,8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
a[0]=a[1]= 1; a[2]=2; a[3]=4; a[n_]:= 6*a[n-2] -a[n-4]; Array[a, 30, 0] (* Robert G. Wilson v, Jun 11 2010 *)
CoefficientList[Series[(1+x-4x^2-2x^3)/((1-2x-x^2)(1+2x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;1;2;4])[1,1] \\ Charles R Greathouse IV, May 10 2016

def A001333(n): return lucas_number2(n, 2, -1)/2
def A006452(n): return (A001333(n+1) + (-1)^n *A001333(n-2))/4
[A006452(n) for n in range(41)] # G. C. Greubel, Jan 22 2023

Bisection: a(2n) = A006452(n). a(2n+1) = A038723(n).
G.f.: ( 1+x-4*x^2-2*x^3 ) / ( (1-2*x-x^2)*(1+2*x-x^2) ).
From Gregory V. Richardson, Oct 07 2002: (Start)
For n (even), a(n) = ( ((3 + sqrt(8))^((n/2)+1) - (3 - sqrt(8))^((n/2)+1)) - 2*((3 + sqrt(8))^((n/2)-1) - (3 - sqrt(8))^((n/2)-1)) ) / (6*sqrt(8)).
For n (odd), a(n) = ( ((3 + sqrt(8))^((n+1)/2) - (3 - sqrt(8))^((n+1)/2)) - 2*((3 + sqrt(8))^((n-1)/2) - (3 - sqrt(8))^((n-1)/2)) ) / (2*sqrt(8)).
Limit_{n->oo} a(n)/a(n-2) = 3 + sqrt(8).
If n is odd, lim_{n->oo} a(n)/a(n-1) = (9 + 2*sqrt(8))/7.
If n is even, lim_{n->oo} a(n)/a(n-1) = (11 + 3*sqrt(8))/7. (End)
a(n+2) = (A001333(n+3) + (-1)^n *A001333(n))/4. - Paul Curtz, Dec 06 2012
a(n+2) = sqrt(17*a(n)^2 + 6*(sqrt(8*a(n)^2 - 7))*a(n)*sgn(2*n - 1) - 7) with a(0) = 1, a(1) = 1. - Raphie Frank, Feb 08 2013
a(n+2) = (A216134(n+2) - A216134(n))/2. - Raphie Frank, Feb 11 2013
E.g.f.: (2*cosh(sqrt(2)*x)*(2*cosh(x) - sinh(x)) + sqrt(2)*(3*cosh(x) - sinh(x))*sinh(sqrt(2)*x))/4. - Stefano Spezia, Nov 26 2022

More terms from James Sellers, May 03 2000

A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

Original entry on oeis.org

0, 3, 15, 120, 528, 4095, 17955, 139128, 609960, 4726275, 20720703, 160554240, 703893960, 5454117903, 23911673955, 185279454480, 812293020528, 6294047334435, 27594051024015, 213812329916328, 937385441796000, 7263325169820735, 31843510970040003, 246739243443988680

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Alternative definition: a(n) is triangular and a(n)/2 is the harmonic average of consecutive triangular numbers. See comments and formula section of A005563, of which this sequence is a subsequence. - Raphie Frank, Sep 28 2012
As with the Sophie Germain triangular numbers (A124174), 35 = (a(n) - a(n-6))/(a(n-2) - a(n-4)). - Raphie Frank, Sep 28 2012
Sophie Germain triangular numbers of the second kind as defined in A217278. - Raphie Frank, Feb 02 2013
Triangular numbers m such that m+1 is a square. - Bruno Berselli, Jul 15 2014
From Vladimir Pletser, Apr 30 2017: (Start)
Numbers a(n) which are the triangular number T(b(n)), where b(n) is the sequence A006451(n) of numbers n such that T(n)+1 is a square.
a(n) also gives the x solutions of the 3rd-degree Diophantine Bachet-Mordell equation y^2 = x^3 + K, with y = T(b(n))*sqrt(T(b(n))+1) = A285955(n) and K = T(b(n))^2 = A285985(n), the square of the triangular number of b(n) = A006451(n).
Also: This sequence is a subsequence of A000217(n), namely A000217(A006451(n)). (End)

			From _Raphie Frank_, Sep 28 2012: (Start)
35*(528 - 15) + 0 = 17955 = a(6),
35*(4095 - 120) + 3 = 139128 = a(7),
35*(17955 - 528) + 15 = 609960 = a(8),
35*(139128 - 4095) + 120 = 4726275 = a(9). (End)
From _Raphie Frank_, Feb 02 2013: (Start)
a(7) = 139128 and a(9) = 4726275.
a(9) = (2*(sqrt(8*a(7) + 1) - 1)/2 + 3*sqrt(a(7) + 1) + 1)^2 - 1 = (2*(sqrt(8*139128 + 1) - 1)/2 + 3*sqrt(139128 + 1) + 1)^2 - 1 = 4726275.
a(9) = 1/2*((3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)^2 + (3*(sqrt(8*a(7) + 1) - 1)/2 + 4*sqrt(a(7) + 1) + 1)) = 1/2*((3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)^2 + (3*(sqrt(8*139128 + 1) - 1)/2 + 4*sqrt(139128 + 1) + 1)) = 4726275. (End)
From _Vladimir Pletser_, Apr 30 2017: (Start)
For n=2, b(n)=5, a(n)=15
For n=5, b(n)=90, a(n)= 4095
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. (End)

Edward J. Barbeau, Pell's Equation, New York: Springer-Verlag, 2003, p. 17, Exercise 1.2.
Allan J. Gottlieb, How four dogs meet in a field, and other problems, Technology Review, Jul/August 1973, pp. 73-74.
Vladimir Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vladimir Pletser, Table of n, a(n) for n = 0..1000 (first 60 terms from Vincenzo Librandi)
M.A. Bennett and A. Ghadermarzi, Data on Mordell's curve.
Michael A. Bennett and Amir Ghadermarzi, Mordell's equation : a classical approach, arXiv:1311.7077 [math.NT], 2013.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jeffrey Shallit, Letter to N. J. A. Sloane, Oct. 1975.
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Eric Weisstein's World of Mathematics, Mordell Curve.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. sqrt(a(n) + 1) = A006452(n + 1) = A216162(2n + 2) and (sqrt(8a(n) + 1) - 1)/2 = A006451.
Cf. A217278, A124174, A216134. - Raphie Frank, Feb 02 2013
Subsequence of A182334.
Cf. A001109, A245031.
Cf. A285955, A285985, A000217, A006451, A081119, A054504. - Vladimir Pletser, Apr 30 2017

I:=[0,3,15,120,528,4095]; [n le 6 select I[n] else 35*(Self(n-2) - Self(n-4)) + Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 21 2015

A006454:=-3*z*(1+4*z+z**2)/(z-1)/(z**2-6*z+1)/(z**2+6*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print ('0,0','1,3','2,15'); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=b*(b+1)/2; print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do: # Vladimir Pletser, Apr 30 2017

Clear[a]; a[0] = a[1] = 1; a[2] = 2; a[3] = 4; a[n_] := 6a[n - 2] - a[n - 4]; Array[a, 40]^2 - 1 (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
LinearRecurrence[{1,34,-34,-1,1},{0,3,15,120,528},30] (* Harvey P. Dale, Feb 18 2023 *)

concat(0, Vec(3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30))) \\ Colin Barker, Apr 30 2017

a(n) = A006451(n)*(A006451(n)+1)/2.
a(n) = A006452(n)^2 - 1. - Joerg Arndt, Mar 04 2011
a(n) = 35*(a(n-2) - a(n-4)) + a(n-6). - Raphie Frank, Sep 28 2012
From Raphie Frank, Feb 01 2013: (Start)
a(0) = 0, a(1) = 3, and a(n+2) = (2x + 3y + 1)^2 - 1  = 1/2*((3x + 4y + 1)^2 + (3x + 4y + 1)) where x = (sqrt(8*a(n) + 1) - 1)/2 = A006451(n) =  1/2*(A216134(n + 1) + A216134(n - 1)) and y = sqrt(a(n) + 1) = A006452(n + 1) = 1/2*(A216134(n + 1) - A216134(n - 1)).
Note that A216134(n + 1) = x + y, and A216134(n + 3) = (2x + 3y + 1) + (3x + 4y + 1) = (5x + 7y + 2), where A216134 gives the indices of the Sophie Germain triangular numbers. (End)
a(n) = (1/64)*(((4 + sqrt(2))*(1 -(-1)^(n+1)*sqrt(2))^(2* floor((n+1)/2)) + (4 - sqrt(2))*(1 + (-1)^(n+1)*sqrt(2))^(2*floor((n+1)/2))))^2 - 1. - Raphie Frank, Dec 20 2015
From Vladimir Pletser, Apr 30 2017: (Start)
Since b(n) = 8*sqrt(T(b(n-2))+1)+ b(n-4) = 8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4), with b(-1)=-1, b(0)=0, b(1)=2, b(2)=5 (see A006451) and a(n) = T(b(n)) (this sequence), we have:
a(n) = ((8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4))*(8*sqrt((b(n-2)*(b(n-2)+1)/2)+1)+ b(n-4)+1)/2). (End)
From Colin Barker, Apr 30 2017: (Start)
G.f.: 3*x*(1 + 4*x + x^2) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) for n > 4.
(End)
a(n)  = (A001109(n/2+1) - 2*A001109(n/2))^2 - 1 if n is even, and (A001109((n+3)/2) - 4*A001109((n+1)/2))^2 - 1 if n is odd (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022

Better description from Harvey P. Dale, Jan 28 2001
More terms from Larry Reeves (larryr(AT)acm.org), Feb 07 2001
Minor edits by N. J. A. Sloane, Oct 24 2009

A045995 Rows of Fibonacci-Pascal triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 8, 3, 1, 1, 5, 55, 55, 5, 1, 1, 8, 610, 6765, 610, 8, 1, 1, 13, 10946, 9227465, 9227465, 10946, 13, 1, 1, 21, 317811, 225851433717, 190392490709135, 225851433717, 317811, 21, 1, 1, 34, 14930352

Offset: 0

N. J. A. Sloane

			1,
1,  1,
1,  1,      1,
1,  2,      2,            1,
1,  3,      8,            3,               1,
1,  5,     55,           55,               5,            1,
1,  8,    610,         6765,             610,            8,      1,
1, 13,  10946,      9227465,         9227465,        10946,     13,  1,
1, 21, 317811, 225851433717, 190392490709135, 225851433717, 317811, 21, 1,
...

Reinhard Zumkeller, Rows n=0..14 of triangle, flattened
R. Whitney, Problem H-254, Fib. Quart., 13 (1975), p. 281.

Cf. A000045, A007318, A006449 (row sums), A081667.

Main diagonal gives A281450.

a045995 n k = a045995_tabl !! n !! k
a045995_row n = a045995_tabl !! n
a045995_tabl = map (map (a000045 . fromInteger)) a007318_tabl
-- Reinhard Zumkeller, Dec 29 2011

A045995 := proc(n,k)
    combinat[fibonacci](binomial(n,k)) ;
end proc: # R. J. Mathar, Dec 03 2014

Flatten[Table[Fibonacci[Binomial[n,k]],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Dec 31 2013 *)

Take Pascal triangle (A007318) and replace each i by Fibonacci(i): a(n,k)=Fibonacci(binomial(n,k)).

More terms from David W. Wilson

cp's OEIS Frontend

A006450 Prime-indexed primes: primes with prime subscripts.

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A006451 Numbers k such that k*(k+1)/2 + 1 is a square.

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A006452 a(n) = 6*a(n-2) - a(n-4).

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A006454 Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.

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A045995 Rows of Fibonacci-Pascal triangle.

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