OFFSET
1,2
COMMENTS
For the initial term, we use A137821(0)=0 (cf. formula).
Sequence A122983 lists record values of this one, which occur at index 2^j (cf. formula). The fact that these values roughly grow by a factor 3 is explained by the fact that these values are given as the sum of all preceding terms (up to +1 or +2 according to the parity of j, cf. formula).
The only values occurring in this sequence are { 1, 2, 3, 7, 8, 21, 61, 62, 183, 547, 548, 1641,... } = A137823, consisting of the record values a(2^j) and, for every other one of these (i.e. for even j), its successor a(2^j)+1, occurring first as a(3*2^j).
The remarkably simple sequence A137824 (= 1,3,2, 4,12,8,...: pattern 1,3,2 multiplied by powers of 4) gives the index at which the value A137823(m) first occurs. - M. F. Hasler, Mar 15 2008
The PARI code given here (function A137822(n)) allows one to calculate hundreds of terms of A107755 in a few microseconds. - M. F. Hasler, Mar 15 2008
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..499.
FORMULA
EXAMPLE
Record values are a(1)=1, a(2)=3, a(4)=7, a(8)=21, a(16)=61, ...
Apart from these values, the only other values occurring in the sequence are:
2=a(1)+1=a(3*1), 8=a(4)+1=a(3*4), 62=a(16)+1=a(3*16), ...
PROG
(PARI) A137822 = D( A137821 ) /* where D(v)=vector(#v-1, i, v[i+1]-v[i]) or D(v)=vecextract(v, "^1")-vecextract(v, "^-1") */
(PARI) n=0; A137822=vector(499, i, { o=n; if( bitand(i, i-1), while(n++ && s+=binomial(4*n-2, 2*n-1)/(2*n)*(10*n-1)/(2*n+1), ), s=Mod(0, 3); n=2*n+1+log(i+.5)\log(2)%2 ); n-o})
(PARI) A137822(n)= local( L=log(n+.5)\log(2) ); while( n>0 || ((n+=2^L) && L=log(n+.5)\log(2)), (n-=2^L) || return( 3^(L+1)\4+1 ); (n-=2^(L-1)) || return( 3^L\4+1+L%2 ); n<0 && n+=2<<L--); 1 \\ M. F. Hasler, Mar 15 2008
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Feb 25 2008, revised Mar 15 2008
STATUS
approved