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A154364 Number of ways to express n as the sum of an odd prime, a positive Pell number and a companion Pell number. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,10

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

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.

EXAMPLE

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

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)<n do for j from 1 while Pell(i)+comp_Pell(j)<n do p:=n-Pell(i)-comp_Pell(j); if (p>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-Fran├žois Alcover, Dec 13 2017, translated from Maple *)

CROSSREFS

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

Sequence in context: A054081 A164585 A200996 * A183043 A050604 A262672

Adjacent sequences:  A154361 A154362 A154363 * A154365 A154366 A154367

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Jan 07 2009

STATUS

approved

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Last modified May 6 15:06 EDT 2021. Contains 343586 sequences. (Running on oeis4.)