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 A292310 Triangular numbers that are equidistant from two other triangular numbers. 4

%I

%S 3,21,28,36,78,105,153,171,190,210,253,325,351,378,465,528,666,703,

%T 903,946,990,1035,1128,1176,1275,1378,1485,1540,1596,1653,1711,1770,

%U 1891,1953,2278,2346,2556,2628,2775,2926,3003,3081,3160,3403,3570,3741,3828,4095,4186,4278,4371,4656

%N Triangular numbers that are equidistant from two other triangular numbers.

%C Triangular numbers which are the arithmetic mean of two other triangular numbers. - _R. J. Mathar_, Oct 01 2017

%F a(n) = A292309(n)/3.

%e 3 is in the sequence because 0 = A000217(0), 6 = A000217(3), and the distances from 3 to 0 and 3 to 6 are the same.

%e 153 is in sequence because 153 = A000217(17), 6 = A000217(2), 300 = A000217(24), and the two distances 300-153 = 153-6 = 147 are the same.

%p isA292310 := proc(n)

%p local ilow ;

%p if isA000217(n) then

%p for ilow from 0 do

%p tilow := A000217(ilow) ;

%p if tilow >= n then

%p return false ;

%p elif isA000217(2*n-tilow) then

%p return true ;

%p end if;

%p end do:

%p else

%p false;

%p end if;

%p end proc:

%p for n from 1 to 5000 do

%p if isA292310(n) then

%p printf("%d,",n) ;

%p end if;

%p end do: # _R. J. Mathar_, Oct 01 2017

%o (PARI) t=3; k=2; while(t<=6000, i=k; e=0; v=t+i; while(i>0&&e==0, if(issquare(8*v+1), e=1; print1(t,", ")); i--; v+=i); k++; t+=k)

%o (PARI) upto(n) = {my(t = 0, i = 0, triangulars = List([0]), res = List); while(t <= n, i++; t+=i; listput(triangulars, t)); for(i=2,#triangulars, tr = triangulars[i]<<1; for(j = 1, i-1, if(issquare(8 * (tr - triangulars[j]) + 1), listput(res, triangulars[i]); next(2)))); res} \\ _David A. Corneth_, Oct 04 2017

%Y Cf. A000217, A292309, A292313, A292314, A292316.

%K nonn

%O 1,1

%A _Antonio RoldÃ¡n_, Sep 14 2017

%E Term 105 added by _David A. Corneth_, Oct 04 2017

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Last modified July 26 13:40 EDT 2021. Contains 346294 sequences. (Running on oeis4.)