A154364 Number of ways to express n as the sum of an odd prime, a positive Pell number and a companion Pell number.

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

Offset: 1

Author

Zhi-Wei Sun, Jan 07 2009

Keywords

• nonn

Comments

This is inspired by the sequence A154290 and related conjectures of Sun. On Jan 08 2009, Zhi-Wei Sun and Qing-Hu Hou conjectured that a(n)>0 for n=6,7,...; in other words, any integer n>5 can be written as the sum of an odd prime, a positive Pell number and a companian Pell number. The Pell numbers are defined by P_0=0, P_1=1 and P_{n+1}=2P_n+P_{n-1} (n=1,2,3,...) and the companion Pell numbers are given by Q_0=Q_1=2 and Q_{n+1}=2Q_n+Q_{n-1} (n=1,2,3...). Note that for n>5 both P_n and Q_n are greater than 2^n.

D. S. McNeil disproved the conjecture by finding the 4 initial counterexamples: 169421772576, 189661491306, 257744272674, 534268276332. - Zhi-Wei Sun, Jan 17 2009

On Feb 01 2009, Zhi-Wei Sun observed that these 4 counterexamples are divisible by 42 and guessed that all counterexamples to the conjecture of Sun and Hou should be multiples of 42. - Zhi-Wei Sun, Feb 01 2009

Examples

			For n=10 the a(10)=3 solutions are 3+5+2, 3+1+6, 7+1+2.
		

Links

• Zhi-Wei Sun, Table of n, a(n), n=1..100000.
• R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
• D. S. McNeil, Sun's strong conjecture, NMBRTHRY, Dec 2008.
• Z. W. Sun, A congruence for primes, Proc. Amer. Math. Soc. 123(1995), 1341-1346.
• Zhi-Wei Sun, A promising conjecture: n=p+F_s+F_t, NMBRTHRY, Dec 2008.
• Zhi-Wei Sun, A summary concerning my conjecture n=p+F_s+F_t, NMBRTHRY, Dec 2008.
• Terence Tao, A remark on primality testing and decimal expansions, arXiv:0802.3361 [math.NT], 2008-2010.
• Terence Tao, A remark on primality testing and decimal expansions, Journal of the Australian Mathematical Society 91:3 (2011), pp. 405-413.

Crossrefs

Cf. A000040, A000129, A002203, A154290, A154257, A154285, A156695.

Programs

Maple

Pell:=proc(n) if n=0 then return(0); elif n=1 then return(1); else return( 2*Pell(n-1) + Pell(n-2) ); fi; end proc: comp_Pell:=proc(n) if n=0 then return(2); elif n=1 then return(2); else return( 2*comp_Pell(n-1) + comp_Pell(n-2) ); fi; end proc: for n from 1 to 10^5 do rep_num:=0; for i from 1 while Pell(i)2) and isprime(p) then rep_num:=rep_num+1; fi; od; od; printf("%d %d\n", n, rep_num); od:

Mathematica

nmax = 10^3;
Pell[n_] := Pell[n] = If[n == 0, Return[0], If[n == 1, Return[1], Return[2* Pell[n - 1] + Pell[n - 2]]]];
compPell[n_] := compPell[n] = If[n == 0, Return[2], If[n == 1, Return[2],  Return[2*compPell[n - 1] + compPell[n - 2]]]];
Reap[For[n = 1, n <= nmax, n++, repnum = 0; For[i = 1, Pell[i] < n, i++, For[j = 1, Pell[i] + compPell[j] < n, j++, p = n - Pell[i] - compPell[j]; If[p > 2 && PrimeQ[p], repnum++]]]; Sow[repnum]]][[2, 1]] (* Jean-François Alcover, Dec 13 2017, translated from Maple *)